%%latex file
%\documentstyle[12pt,amssymb]{amsart}
\documentclass[12pt]{amsart} \usepackage{amssymb} \usepackage{mathabx} \usepackage{bm} \usepackage{wasysym} \usepackage{graphicx} \usepackage{stmaryrd}
\usepackage{imakeidx}
\usepackage{color,hyperref}
\hypersetup{colorlinks,breaklinks,
             linkcolor=blue,urlcolor=blue,
           anchorcolor=blue,citecolor=blue}
\newenvironment{pf}{\proof[\proofname]}{\endproof}

\makeindex[name=terms,title=Index of terms, columns=2]

\renewcommand\baselinestretch{1.1}
\renewcommand\theequation{\arabic{section}.\arabic{equation}}

\newcommand{\wrlab}[1]{\label{#1}}
\renewcommand{\arraystretch}{1}

\newtheorem{thm}{THEOREM}[section]
\numberwithin{thm}{subsection}
\newtheorem{axiom}[thm]{AXIOM}
\newtheorem{defn}[thm]{DEFINITION}

\newcommand{\secref}[1]{\S\ref{#1}}

\newcommand{\tref}[1]{Theorem~\ref{#1}}
\newcommand{\aref}[1]{Axiom~\ref{#1}}
\newcommand{\dref}[1]{Definition~\ref{#1}}

% BLACK BOARD BOLD

\def\Bbb{\mathbb}  %needed in Nancy

\def\A{{\mathbb A}} \def\B{{\mathbb B}} \def\C{{\mathbb C}} \def\D{{\mathbb D}}
\def\N{{\mathbb N}} \def\I{{\mathbb I}} \def\Q{{\mathbb Q}} \def\R{{\mathbb R}}
\def\Z{{\mathbb Z}} 

\def\T{{\mathbb T}}

\def\P{{\mathbb P}} 

% BOLD
\def\bfzero{{\bf 0}}
\def\bfP{{\bf P}}
\def\bfQ{{\bf Q}}
\def\bfS{{\bf S}}
\def\bfT{{\bf T}}

% CAL 
\def\Cal{\mathcal} \def\scra{{\mathcal A}} \def\scrb{{\mathcal B}}
\def\scrd{{\mathcal D}} \def\scrf{{\mathcal F}} \def\scrg{{\mathcal G}}
\def\scri{{\mathcal I}} \def\scrk{{\mathcal K}} \def\scrl{{\mathcal L}}
\def\scrm{{\mathcal M}} \def\scrp{{\mathcal P}} \def\scrq{{\mathcal Q}}
\def\scrr{{\mathcal R}} \def\scrs{{\mathcal S}} \def\scrt{{\mathcal T}}
\def\scru{{\mathcal U}} \def\scrv{{\mathcal V}} \def\scrw{{\mathcal W}}
\def\scre{{\mathcal E}} \def\scrx{{\mathcal X}} \def\scry{{\mathcal Y}}
\def\scrz{{\mathcal Z}} \def\scrh{{\mathcal H}} \def\scrc{{\mathcal C}}
\def\scrn{{\mathcal N}} \def\scro{{\mathcal O}}
\def\scrpf{{\scrp\scrf}}
\def\scroi{{\scro\scri}}

% HATTED ROMAN
\def\hatd{{\widehat{d}}}
\def\hatX{{\widehat{X}}}

% HATTEDCAL
\def\hatscrp{{\widehat{\scrp}}}
\def\hatscrp{{\widehat{\scrp}}}
\def\hatpi{{\widehat{\pi}}}

% HATTED BB BOLD
\def\hatbbR{{\widehat{\R}}}

\newcommand\scrlito{
  \mathchoice
    {{\scriptstyle\mathcal{O}}}% \displaystyle
    {{\scriptstyle\mathcal{O}}}% \textstyle
    {{\scriptscriptstyle\mathcal{O}}}% \scriptstyle
    {\scalebox{.7}{$\scriptscriptstyle\mathcal{O}$}}%\scriptscriptstyle
  }

\def\scrbigo{{\widehat{\scro}}}

% LIE ALGEBRAS

\font\tenolde=eufm10 at 10pt
\font\sevenolde=eufm7
\font\fiveolde=eufm5
\newfam\oldefam
\textfont\oldefam=\tenolde
\scriptfont\oldefam=\sevenolde
\scriptscriptfont\oldefam=\fiveolde
\def\fraktwo{\fam\oldefam\tenolde}

\newfam\cyrfam
\font\tencyr=wncyr10
\def\cyr{\fam\cyrfam\tencyr}

\newcommand\sbullet[1][.75]{\mathbin{\vcenter{\hbox{\scalebox{#1}{$\bullet$}}}}}

\def\aff{{\mathfrak{aff}(\R)}} \def\La{{\mathfrak a}}
\def\Lb{{\mathfrak b}} \def\Lc{{\mathfrak c}} \def\Lg{{\mathfrak g}}
\def\Lh{{\mathfrak h}} \def\Lk{{\mathfrak k}} \def\Ll{{\mathfrak l}}
\def\Lm{{\mathfrak m}} \def\Ln{{\mathfrak n}} \def\Lr{{\mathfrak r}}
\def\Ls{{\mathfrak s}} \def\Lt{{\mathfrak t}} \def\Lw{{\mathfrak w}}
\def\Lq{{\mathfrak q}} \def\Lu{{\mathfrak u}} \def\Lv{{\mathfrak v}}
\def\Lz{{\mathfrak z}} \def\Lsl{\mathfrak{sl}} \def\Lgl{\mathfrak{gl}}
\def\Lp{{\mathfrak p}} \def\Lf{{\mathfrak f}} \def\Lx{{\mathfrak x}}
\def\Lpsl{\mathfrak{psl}}
\def\Lso{{\mathfrak{so}}} 
\def\Lsu{{\mathfrak{su}}} 
\def\Lsp{{\mathfrak{sp}}} 
\def\orthog{{\mathfrak{o}}} 
\def\tilso{{\mathfrak{so}}}
\def\tilco{{\mathfrak{co}}} 
\def\co{\mathfrak{co}}

% OVERLINE
\def\barscru{{\overline{\scru}}}
\def\barPsi{{\overline{\Psi}}}
\def\barA{{\overline{A}}}
\def\barB{{\overline{B}}}
\def\barC{{\overline{C}}}
\def\barD{{\overline{D}}}
\def\barE{{\overline{E}}}
\def\barF{{\overline{F}}}
\def\barbarF{{\overline{\overline{F}}}}
\def\barG{{\overline{G}}}
\def\barH{{\overline{H}}}
\def\barI{{\overline{I}}}
\def\barJ{{\overline{J}}}
\def\barK{{\overline{K}}}
\def\barL{{\overline{L}}}
\def\barM{{\overline{M}}}
\def\barN{{\overline{N}}}
\def\barP{{\overline{P}}}
\def\barbarP{{\overline{\overline{P}}}}
\def\barQ{{\overline{Q}}}
\def\barR{{\overline{R}}}
\def\barS{{\overline{S}}}
\def\barT{{\overline{T}}}
\def\barU{{\overline{U}}}
\def\barV{{\overline{V}}}
\def\barW{{\overline{W}}}
\def\barbarW{{\overline{\overline{W}}}}
\def\barX{{\overline{X}}}
\def\barY{{\overline{Y}}}
\def\barZ{{\overline{Z}}}
\def\barLr{{\overline{\Lr}}}
\def\barLg{{\overline{\Lg}}}
\def\barLh{{\overline{\Lh}}}
\def\barLp{{\overline{\Lp}}}
\def\barLn{{\overline{\Ln}}}
\def\barLm{{\overline{\Lm}}}
\def\barLl{{\overline{\Ll}}}
\def\bara{{\overline{a}}}
\def\bare{{\overline{e}}}
\def\barf{{\overline{f}}}
\def\barg{{\overline{g}}}
\def\bark{{\overline{k}}}
\def\baru{{\overline{U}}}
\def\barv{{\overline{v}}}
\def\barw{{\overline{w}}}
\def\barz{{\overline{z}}}
\def\barbeta{{\overline{\beta}}}
\def\bargamma{{\overline{\gamma}}}
\def\barI{{\overline{I}}}

% TILDE
\def\tilvec{\widetilde{\vec}}

% HAT
\def\hatA{{\widehat{A}}}
\def\hatC{{\widehat{C}}}
\def\hatH{{\widehat{H}}}
\def\hatI{{\widehat{I}}}
\def\hatN{{\widehat{N}}}
\def\hatQ{{\widehat{Q}}}
\def\hatR{{\widehat{R}}}
\def\hatS{{\widehat{S}}}
\def\hatV{{\widehat{V}}}
\def\hatt{{\widehat{t}}}
\def\hatu{{\widehat{u}}}
\def\hatv{{\widehat{v}}}
\def\hatw{{\widehat{w}}}
\def\hatx{{\widehat{x}}}
\def\haty{{\widehat{y}}}
\def\hatz{{\widehat{z}}}
\def\hatalpha{{\widehat{\alpha}}}
\def\hatrho{{\widehat{\rho}}}
\def\hatkappa{{\widehat{\kappa}}}
\def\hatlambda{{\widehat{\lambda}}}
\def\hatsigma{{\widehat{\sigma}}}
\def\hatmu{{\widehat{\mu}}}
\def\hatnu{{\widehat{\nu}}}
\def\hattau{{\widehat{\tau}}}

% CHECK
\def\ckt{{\widecheck{t}}}
\def\ckw{{\widecheck{w}}}
\def\ckx{{\widecheck{x}}}
\def\cky{{\widecheck{y}}}
\def\ckz{{\widecheck{z}}}
\def\cksigma{{\widecheck{\sigma}}}
\def\cktau{{\widecheck{\tau}}}
\def\ckH{{\widecheck{H}}}
\def\ckN{{\widecheck{N}}}
\def\ckS{{\widecheck{S}}}

\def\ds{\displaystyle}
\def\bfindex[#1]{{\bf#1}\index[terms]{#1}}

\def\timestimes{{\times\!\!\times}}

%operators
\def\TJOC{{\rm TJOC}}
\def\JOC{{\rm JOC}}
\def\PF{{\rm PF}}
\def\set{{\rm set}}
\def\bi{{\rm bi}}
\def\fn{{\rm fn}}
\def\ss{{\rm setset}}
\def\sf{{\rm setfn}}
\def\fs{{\rm fnset}}
\def\seq{{\rm seq}}
\def\EUCL{{\rm EUCL}}
\def\TNSR{{\rm TNSR}}
\def\starstar{{**}}
\def\fincap{{\hbox{{\rm fin}}\,\cap}}
\def\fincup{{\hbox{{\rm fin}}\,\cup}}
\def\ctblcap{{\hbox{{\rm ctbl}}\,\cap}}
\def\ctblcup{{\hbox{{\rm ctbl}}\,\cup}}
\def\finamalg{{\hbox{{\rm fin}}\,\amalg}}
\def\two{{\rm two}}
\def\three{{\rm three}}
\def\ten{{\rm ten}}
\def\Lin{{\rm Lin}}
\def\Bilin{{\rm Bilin}}
\def\Trilin{{\rm Trilin}}
\def\PS{{\rm PS}}
\def\UE{{\rm UE}}
\def\FV{{\rm FV}}
\def\CH{{\rm CH}}
\def\DQ{{\rm DQ}}
\def\SS{{\rm SS}}
\def\SUMS{{\rm SUMS}}
\def\res{{\rm res}}
\def\add{{\rm add\,}}
\def\FS{{\rm FS}}
\def\LIMS{{\rm LIMS}}
\def\NOT{{\rm NOT}}
\def\ORB{{\rm ORB}}
\def\dlin{{\rm dlin}}
\def\LINS{{\rm LINS}}
\def\opLIMS{\mathop{\LIMS}}
\def\IDS{{\rm IDS}}
\def\UL{{\rm UL}}
\def\UB{{\rm UB}}
\def\LB{{\rm LB}}
\def\LP{{\rm LP}}
\def\LPD{{\rm LPD}}
\def\Isol{{\rm Isol}}
\def\Cl{{\rm Cl}}
\def\sCl{{\rm sCl}}
\def\id{{\rm id}}
\def\dim{{\rm dim\,}}
\def\DIMS{{\rm DIMS\,}}
\def\Bas{{\rm Bas\,}}
\def\Int{{\rm Int}}
\def\IntD{{\rm IntD}}
\def\det{{\rm det\,}}
\def\exp{{\rm exp}}
\def\End{{\rm End}}
\def\SO{{\rm SO}}
\def\CO{{\rm CO}}
\def\O{{\rm O}}
\def\Stab{{\rm Stab}}
\def\Fix{{\rm Fix}}
\def\GL{{\rm GL}}
\def\Ad{{\rm Ad}}
\def\Aut{{\rm Aut}}
\def\der{{\rm der}}
\def\ad{{\rm ad}}
\def\diag{{\rm diag}}
\def\Mink{{\rm Mink}}
\def\Tay{{\rm Tay}}
\def\lin{{\rm lin}}
\def\switch{{\rm switch}}
\def\SOtwortwo{{\SO(2)\ltimes\R^2}}
\def\O{{\rm O}}
\def\OB{{\rm OB}}
\def\SO{{\rm SO}}
\def\SL{{\rm SL}}
\def\PSL{{\rm PSL}}
\def\SBF{{\rm SBF}}
\def\STF{{\rm STF}}
\def\SM{{\rm SM}}
\def\QF{{\rm QF}}
\def\Id{{\rm Id}}
\def\dom{{\rm dom}}
\def\im{{\rm im}}
\def\dd{{\rm dd}}
\def\dct{{\rm dct}}
\def\sym{{\rm sym}}
\def\Sym{{\rm Sym}}
\def\as{{\rm as}}
\def\mm{{\rm mm}}
\def\cof{{\rm cof}}
\def\adj{{\rm adj}}
\def\det{{\rm det}}
\def\DET{{\rm DET}}
\def\ADJ{{\rm ADJ}}
\def\DNZ{{\hbox{\rm DNZ}}}
\def\CVZ{{\hbox{\rm CVZ}}}
\def\BNZ{{\hbox{\rm BNZ}}}
\def\ev{{\rm ev}}
\def\sl{{\rm s\ell}}
\def\JO{{\rm JO}}
\def\TL{{\rm TL}}
\def\supp{{\rm supp}}

%misc
\def\vep{{\varepsilon}}
\def\bksl{{\backslash}}
\def\sigmaalg{{\rm s.alg}}
\def\drta{\dashrightarrow}
\def\ltlt{{<\!\!<}}
\def\phx{\phantom{x}}

\def\Xint#1{\mathchoice
{\XXint\displaystyle\textstyle{#1}}%
{\XXint\textstyle\scriptstyle{#1}}%
{\XXint\scriptstyle\scriptscriptstyle{#1}}%
{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
\!\int}
\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$ }
\vcenter{\hbox{$#2#3$ }}\kern-.6\wd0}}
\def\ddashint{\Xint=}
\def\dashint{\Xint-}

\begin{document}

\centerline{\bf Solutions for MATH 4603 (Advanced Calculus I)}
\centerline{\bf Fall 2022}
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 13: Due on Tuesday 14 December}
\vskip.1in\hrule\vskip.1in\noindent
13-1. Show: Let $S,T\subseteq\R$. \quad Then $\JO_{S\cup T}\,\,\le\,\,\JO_S\,+\,\JO_T$.
\vskip.1in\noindent
{\it Proof:}\\
Let $Q:=\{\,\TL_Z\,|\,(\,Z\in\scroi^\N\,)\,\&\,(\,\bigcup\I_Z\supseteq S\bigcup T\,)\,\&\,(\,\#\supp\,Z<\infty\,)\,\}$.\\
Let $A:=\{\,\TL_U\,|\,(\,U\in\scroi^\N\,)\,\&\,(\,\bigcup\I_U\supseteq S\,)\,\&\,(\,\#\supp\,U<\infty\,)\,\}$.\\
Let $B:=\{\,\TL_V\,|\,(\,V\in\scroi^\N\,)\,\&\,(\,\bigcup\I_V\supseteq T\,)\,\&\,(\,\#\supp\,V<\infty\,)\,\}$.\\
Then $\,$ $\JO_{S\cup T}=\inf_Q$ $\,$ and $\,$ $\JO_S=\inf_A$ $\,$ and $\,$ $\JO_T=\inf_B$.\\
{\bf Want:} $\inf_Q\,\le\,\inf_A\,+\,\inf_B$.\\
Exactly one of the following is true:\\
\phx\qquad (1) $\inf_A+\inf_B=\infty$ \qquad or \qquad (2) $\inf_A+\inf_B\ne\infty$.\\
\\
{\it Case (1):}\\
We have $\inf_Q\in\R^*\le\infty=\inf_A+\inf_B$.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
We have $\inf_A=\JO_S\ge0$ and $\inf_B=\JO_T\ge0$,\\
\phx\qquad so $\inf_A\ge0$ and $\inf_B\ge0$.\\
Then $\inf_A\le\inf_A+\inf_B$ and $\inf_B\le\inf_A+\inf_B$.\\
Also, $\inf_A+\inf_B\ge0$, so $\inf_A+\inf_B\in[0;\infty]$.\\
So, since $\inf_A+\inf_B\ne\infty$, we get: $\inf_A+\inf_B\in[0;\infty]\backslash\{\infty\}$.\\
Then $\inf_A+\inf_B\in[0;\infty]\backslash\{\infty\}=[0;\infty)<\infty$,\\
\phx\qquad  so $\inf_A+\inf_B<\infty$.\\
Then $\inf_A\le\inf_A+\inf_B<\infty$ and $\inf_B\le\inf_A+\inf_B<\infty$,\\
\phx\qquad so $\inf_A<\infty$ and $\inf_B<\infty$.\\
{\bf Want:} $\forall\vep>0$, $\,$ $\inf_Q\,\le\,\inf_A\,+\,\inf_B+\vep$.\\
Given $\vep>0$. \quad {\bf Want:} $\inf_Q\,\le\,\inf_A\,+\,\inf_B+\vep$.\\
We have $\inf_A\ge0>-\infty$ and $\inf_A<\infty$.\\
Then $-\infty<\inf_A<\infty$, so $\inf_A+(\vep/2)>\inf_A$.\\
Since $\neg(\inf_A+(\vep/2)\le\inf_A)$, we get $\neg(\inf_A+(\vep/2)\le A)$.\\
Since $\neg(\inf_A+(\vep/2)\le A)$, choose $a\in A$ s.t.~$\inf_A+(\vep/2)>a$.\\
We have $\inf_B\ge0>-\infty$ and $\inf_B<\infty$.\\
Then $-\infty<\inf_B<\infty$, so $\inf_B+(\vep/2)>\inf_B$.\\
Since $\neg(\inf_B+(\vep/2)\le\inf_B)$, we get $\neg(\inf_B+(\vep/2)\le B)$.\\
Since $\neg(\inf_B+(\vep/2)\le B)$, choose $b\in B$ s.t.~$\inf_B+(\vep/2)>b$.\\
Since $a\in A=\{\,\TL_U\,|\,(\,U\in\scroi^\N\,)\,\&\,(\,\bigcup\I_U\supseteq S\,)\,\&\,(\,\#\supp\,U<\infty\,)\,\}$,\\
\phx\quad choose $U\in\scroi^\N$ s.t.~$\bigcup\I_U\supseteq S$ and $\#\supp\,U<\infty$ and $\TL_U=a$.\\
Since $b\in B=\{\,\TL_V\,|\,(\,V\in\scroi^\N\,)\,\&\,(\,\bigcup\I_V\supseteq T\,)\,\&\,(\,\#\supp\,V<\infty\,)\,\}$,\\
\phx\quad choose $V\in\scroi^\N$ s.t.~$\bigcup\I_V\supseteq T$ and $\#\supp\,V<\infty$ and $\TL_V=b$.\\
Let $Z:=(U_1,V_1,U_2,V_2,U_3,V_3,\ldots)$. \quad Then $Z\in\scroi^\N$.\\
Then, by a class theorem, we have:\\
\phx\qquad\quad $\I_Z=\I_U\bigcup\I_V$ \quad and \quad $\#\supp\,Z=\#\supp\,U+\#\supp\,V$\\
\phx\qquad\qquad\quad and \quad $\TL_Z=\TL_U+\TL_V$.\\
Since $\supp\,U<\infty$ and $\supp\,V<\infty$, we get: $\#\supp\,U+\#\supp\,V<\infty$.\\
Then $\#\supp\,Z=\#\supp\,U+\#\supp\,V<\infty$, \quad so $\#\supp\,Z<\infty$.\\
Since $\I_U\supseteq S$ and $\I_V\supseteq T$, we get: $\I_U\bigcup\I_V\supseteq S\bigcup T$.\\
Then $\bigcup\I_Z=\bigcup\big(\I_U\bigcup\I_V\big)=\big(\bigcup\I_U\big)\bigcup\big(\bigcup\I_V\big)\supseteq S\bigcup T$,\\
\phx\quad so $\bigcup\I_Z\supseteq S\bigcup T$.\\
Since $Z\in\scroi^\N$ and $\bigcup\I_Z\supseteq S\bigcup T$ and $\#\supp\,Z<\infty$,  we get:\\
\phx\quad $\TL_Z\in\{\,\TL_Z\,|\,(\,Z\in\scroi^\N\,)\,\&\,(\,\bigcup\I_Z\supseteq S\bigcup T\,)\,\&\,(\,\#\supp\,Z<\infty\,)\,\}$.\\
So, by definition of $Q$, we get $\TL_Z\in Q$.\\
Then $a+b=\TL_U+\TL_V=\TL_Z\in Q\ge\inf_Q$, so $\inf_Q\le a+b$.\\
Then $\inf_Q\le a+b<(\inf_A+(\vep/2))+(\inf_B+(\vep/2))$.\\
Then $\inf_Q<\inf_A+\inf_B+\vep$, so $\inf_Q\le\inf_A+\inf_B+\vep$.\\
{\it End of Case (2).} \quad 
QED
\vskip.1in\hrule\vskip.1in\noindent
13-2. Show: $\forall k\in\N$,\\
\phx\qquad\qquad\qquad $\forall S_1,\ldots,S_k\subseteq\R$, \quad $\JO_{S_1\cup\cdots\cup S_k}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_k}$.
\vskip.1in\noindent
{\it Proof:}\\
Let $A:=\{k\in\N\,|\,\forall S_1,\ldots,S_k\subseteq\R,\,\JO_{S_1\cup\cdots\cup S_k}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_k}\}$.
We have: $\forall S_1\subseteq\R$, $\,$ $\JO_{S_1}\le\JO_{S_1}$. \quad Then $1\in A$.\\
By the PMI, want: $\forall k\in A$, $k+1\in A$.\\
Given $k\in A$. \quad {\bf Want:} $k+1\in A$.\\
Know: $\forall S_1,\ldots,S_k\subseteq\R$, \quad $\JO_{S_1\cup\cdots\cup S_k}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_k}$.\\
{\bf Want:} $\forall S_1,\ldots,S_{k+1}\subseteq\R$, \quad $\JO_{S_1\cup\cdots\cup S_{k+1}}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_{k+1}}$.\\
Given $S_1,\ldots,S_{k+1}\subseteq\R$. {\bf Want:} $\JO_{S_1\cup\cdots\cup S_{k+1}}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_{k+1}}$.\\
Know: $\JO_{S_1\cup\cdots\cup S_k}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_k}$.\\
Then $\JO_{S_1\cup\cdots\cup S_k}+\JO_{S_{k+1}}\,\,\le\,\,(\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_k}\,)\,+\,\JO_{S_{k+1}}$.\\
Then $\JO_{S_1\cup\cdots\cup S_k}+\JO_{S_{k+1}}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_{k+1}}$.\\
By a class theorem, $\JO_{(S_1\cup\cdots\cup S_k)\cup S_{k+1}}\,\,=\,\,\JO_{S_1\cup\cdots\cup S_k}+\JO_{S_{k+1}}$.\\
Then $\JO_{S_1\cup\cdots\cup S_{k+1}}\,\,=\,\,\JO_{S_1\cup\cdots\cup S_k}+\JO_{S_{k+1}}$.\\
Then $\JO_{S_1\cup\cdots\cup S_{k+1}}\,\,=\,\,\JO_{S_1\cup\cdots\cup S_k}+\JO_{S_{k+1}}\,\,\le\,\,\JO_{S_1}\,+\,\cdots\,+\,\JO_{S_{k+1}}$.\\
QED
\vskip.1in\hrule\vskip.1in\noindent
13-3. Show: Let $k\in\N$. \quad
Assume: $\,$ $\forall\alpha\in\R$, $\forall\beta\ge\alpha$, $\,$ $\JO_{[\alpha;\beta]}^k\ge\beta-\alpha$.\\
Let $a\in\R$. $\,$ Let $b\ge a$. $\,$ Let $I:=[a;b]$. $\,$ Then $\JO_I^{k+1}\ge b-a$.
\vskip.1in\noindent
{\it Proof:}
We have: \quad $\forall\alpha\in\R$, $\forall\beta<\alpha$, $\,$ $\JO_{[\alpha;\beta]}^k=\JO_\emptyset^k=0>\beta-\alpha$.\\
Then: \quad $\forall\alpha,\beta\in\R$, $\,$ $\JO_{[\alpha;\beta]}^k\ge\beta-\alpha$.\\
{\bf Want:} $\inf(\TJOC_I^{k+1})\ge b-a$. \qquad
{\bf Want:} $\TJOC_I^{k+1}\ge b-a$.\\
Assume: $\neg(\TJOC_I^{k+1}\ge b-a)$. \quad {\bf Want:} Contradiction.\\
Choose $v\in\TJOC_I^{k+1}$ s.t.~$v<b-a$.\\
Choose $U\in\JOC_I^{k+1}$ s.t.~$v=TL_U$.\\
Since $U\in\JOC_I^{k+1}$, we know:\\
\phx\quad $U\in\scro\scri^\N$ \quad and \quad $\#\supp\,U\le k+1$ \quad and \quad $\bigcup\I_U\supseteq I$.\\
Since $b\in[a;b]=I\subseteq\bigcup\I_U$, choose $W\in\I_U$ s.t.~(if you wish)$b\in W$.\\
Since $W\in\I_U$ and since $U\in\scro\scri^\N$, choose $j\in\N$ s.t.~$W=U_j$.\\
Since $b\in W=U_j$, we get: $b\in U_j$. \quad Then $U_j\ne\emptyset$.\\
Since $U_j\in\scro\scri$, choose $\beta,\gamma\in\R$ s.t.~$U_j=(\beta;\gamma)$.\\
Then, since $U_j\ne\emptyset$, we get: $\beta<\gamma$. \quad Then $L_{U_j}=\gamma-\beta$.\\
Since $b\in U_j=(\beta;\gamma)$, we get: $\beta<b<\gamma$.\\
Then $I\backslash U_j=[a;b]\backslash(\beta;\gamma)=[a;\beta]$.\\
Let $\alpha:=a$. \quad Then $\JO_{[\alpha;\beta]}^k\ge\beta-\alpha$. \quad Also, $I\backslash U_j=[\alpha;\beta]$.\\
Let $J:=[\alpha;\beta]$. \quad Then $\JO_J^k\ge\beta-\alpha$. \quad Also, $I\backslash U_j=J$.\\
\vskip-0.15in\noindent
Define $V\in\scro\scri^\N$ by: \quad $\forall i\in\N$, $\,$ $V_i=\begin{cases}U_i,&\hbox{if }i\ne j\\\emptyset,&\hbox{if }i=j.\end{cases}$\\
\vskip-0in\noindent
Then: \quad $V\in\scro\scri^\N$ \quad and \quad $\#\supp\,V\le k$ \quad and\\
\phx\qquad\qquad\quad $\bigcup\I_V\supseteq(\bigcup\I_U)\backslash U_j$ \quad and \quad $\TL_V=\TL_U-L_{U_j}$.\\
We have: \quad $\,\bigcup\I_V\supseteq(\bigcup\I_U)\backslash U_j\supseteq I\backslash U_j=J$.\\
Then $V\in\JOC_J^k$. \quad Then $\TL_V\in\TJOC_J^k$.\\
Then $\TL_V\ge\inf(\TJOC_J^k)=\JO_J^k\ge\beta-\alpha$, \quad so $\TL_V\ge\beta-\alpha$.\\
Then $v=\TL_U=\TL_V+L_{U_j}=\TL_V+(\gamma-\beta)\ge(\beta-\alpha)+(\gamma-\beta)$.\\
Then $v\ge\gamma-\alpha$. \quad Recall: $b<\gamma$ and $a=\alpha$.\\
Then $v\ge\gamma-\alpha>b-\alpha=b-a$, \quad so $v>b-a$.\\
However, by choice of $v$, we have: $v<b-a$. \quad Contradiction. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
13-4. Show: Let $T\subseteq\R$. \quad Then $\Int\,T$ is open.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall a\in\Int\,T$, $\exists B\in\scrb(x)$ s.t.~$B\subseteq\Int\,T$.\\
Given $a\in\Int\,T$. {\bf Want:} $\exists B\in\scrb(x)$ s.t.~$B\subseteq\Int\,T$.\\
Since $a\in\Int\,T$, choose $B\in\scrb(x)$ s.t.~$B\subseteq T$.\\
Then $B\in\scrb(x)$. \quad {\bf Want:} $B\subseteq\Int\,T$.\\
Since $B\in\scrb(x)\subseteq\scrb_\R$, by a class theorem,\\
\phx\quad we conclude that $B$ is open.\\
Then $B=\Int\,B$.\\
Since $B\subseteq T$, by a class theorem,\\
\phx\quad we conclude that $\Int\,B\subseteq\Int\,T$.\\
Then $B=\Int\,B\subseteq\Int\,T$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
13-5. Show: Let $S\subseteq\R$. \quad Then $\Cl\,S$ is closed.
\vskip.1in\noindent
{\it Proof:}
Let $T:=\R\backslash S$.\\
By a class theorem $\partial T=\partial S$.\\
Then $\Int\,T=T\backslash\partial T=T\backslash\partial S\subseteq(\R\backslash S)\backslash\partial S\subseteq\R\backslash(S\cup\partial S)=\R\backslash(\Cl\,S)$,\\
\phx\quad so $\Int\,T=\R\backslash(\Cl\,S)$, \quad so $\R\backslash(\Int\,T)=\Cl\,S$.\\
{\bf Want:} $\R\backslash(\Int\,T)$ is closed.\\
By HW\#13-4, we get: $\Int\,T$ is open.\\
Then, by a class theorem, $\R\backslash(\Int\,T)$ is closed. \quad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 12: Due on Tuesday 6 December}
\vskip.1in\hrule\vskip.1in\noindent
12-1. Using the PMI, show: Let $a\in\R$. Then:\\
\phx\qquad\qquad $\forall j\in\N$, \qquad $((\bullet)^j)^\T_a-j\cdot a^{j-1}\cdot(\bullet)\in\scrlito(1)$.
\vskip.1in\noindent
{\it Proof:}
Let $I:=\{j\in\N\,|\,((\bullet)^j)^\T_a-j\cdot a^{j-1}\cdot(\bullet)\in\scrlito(1)\}$.
{\bf Want:} $I=\N$.\\
We have: $((\bullet)^1)^\T_a-1\cdot a^{1-1}\cdot(\bullet)=((\bullet))^\T_a-1\cdot 1\cdot(\bullet)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $=(\bullet)-(\bullet)=\bfzero\in\scrlito(1)$. \qquad\quad
Then $1\in I$.\\
By the PMI, want: $\forall j\in I$, $j+1\in I$.\\
Given $j\in I$. \quad {\bf Want:} $j+1\in I$.\\
Know: $((\bullet)^j)^\T_a-j\cdot a^{j-1}\cdot(\bullet)\in\scrlito(1)$.\\
{\bf Want:} $((\bullet)^{j+1})^\T_a-(j+1)\cdot a^{(j+1)-1}\cdot(\bullet)\in\scrlito(1)$.\\
Let $R:=((\bullet)^j)^\T_a-j\cdot a^{j-1}\cdot(\bullet)$.\\
Let $S:=(\bullet)^{j+1})^\T_a-(j+1)\cdot a^{(j+1)-1}\cdot(\bullet)$.\\
Know: $R\in\scrlito(1)$. \quad {\bf Want:} $S\in\scrlito(1)$.\\
We have: $\forall h\in\R$,\\
\phx\qquad\qquad\qquad\qquad $\,\,\,R_h=(a+h)^j-a^j-j\cdot a^{j-1}\cdot h$,\\
\phx\qquad and so \quad $(a+h)^j=a^j+j\cdot a^{j-1}\cdot h+R_h$.\\
We have: $\forall h\in\R$,\\
\phx\qquad $S_h=(a+h)^{j+1}-a^{j+1}-(j+1)\cdot a^{(j+1)-1}\cdot h$\\
\phx\qquad\quad $=(a+h)\cdot(a+h)^j-a^{j+1}-(j+1)\cdot a^j\cdot h$\\
\phx\qquad\quad $=(a+h)\cdot(a^j+j\cdot a^{j-1}\cdot h+R_h)-a^{j+1}-(j+1)\cdot a^j\cdot h$\\
\phx\qquad\quad $=\quad\,\, a\cdot a^j+a\cdot j\cdot a^{j-1}\cdot h+a\cdot R_h$\\
\phx\qquad\qquad $\,\,+\,h\cdot a^j+h\cdot j\cdot a^{j-1}\cdot h+h\cdot R_h$\\
\phx\qquad\qquad\qquad\qquad $-\,a^{j+1}-(j+1)\cdot a^j\cdot h$\\
\phx\qquad\quad $=\quad a^{j+1}+\qquad\quad\, j\cdot a^j\cdot h+a\cdot R_h$\\
\phx\qquad\qquad\qquad $\,\,\,\,\,\,\,+\qquad\qquad\,\,\, a^j\cdot h+\qquad\qquad j\cdot a^{j-1}\cdot h^2+h\cdot R_h$\\
\phx\qquad\qquad $\,\,-\,a^{j+1}-\,\,(j+1)\cdot a^j\cdot h$\\
\phx\qquad\quad $=\qquad\qquad\qquad\qquad\qquad\qquad a\cdot R_h+j\cdot a^{j-1}\cdot h^2+h\cdot R_h$\\
\phx\qquad\quad $=(a\cdot R+j\cdot a^{j-1}\cdot(\bullet)^2+(\bullet)\cdot R)_h$.\\
Then $S=a\cdot R+j\cdot a^{j-1}\cdot(\bullet)^2+(\bullet)\cdot R$.\\
By algebraic linearity of $\scrlito(1)$, want: $a\cdot R\,,\,j\cdot a^{j-1}\cdot(\bullet)^2\,,\,(\bullet)\cdot R\,\in\,\scrlito(1)$.\\
Since $R\in\scrlito(1)$, by algebraic linearaity of $\scrlito(1)$,\\
\phx\qquad we conclude: $a\cdot R\,\in\,\scrlito(1)$.\\
We have $j\cdot a^{j-1}\cdot(\bullet)^2\,\in\,\scrq\,=\,\scrh_2\,\subseteq\,\scrbigo(2)\,\subseteq\,\scrlito(1)$.\\
{\bf Want:} $(\bullet)\cdot R\,\in\,\scrlito(1)$.\\
Since $(\bullet)\in\scrl=\scrh_1\subseteq\scrbigo(1)$ and $R\in\scrlito(1)$,\\
\phx\qquad we conclude: $(\bullet)\cdot R\,\in\,\scrlito(1+1)$.\\
Then $(\bullet)\cdot R\,\in\,\scrlito(1+1)\,=\,\scrlito(2)\,\subseteq\,\scrlito(1)$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
12-2. Show: \quad Let $j\in\N$. \quad Then $((\bullet)^j)'=j\cdot(\bullet)^{j-1}$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall a\in\R$, $((\bullet)^j)'_a=(j\cdot(\bullet)^{j-1})_a$.\\
Given $a\in\R$. \quad {\bf Want:} $((\bullet)^j)'_a=(j\cdot(\bullet)^{j-1})_a$.\\
{\bf Want:} $((\bullet)^j)'_a=j\cdot a^{j-1}$. \quad
By HW\#12-1, $((\bullet)^j)^\T_a-j\cdot a^{j-1}\cdot(\bullet)\in\scrlito(1)$.\\
Then $j\cdot a^{j-1}\cdot(\bullet)\,\,\in\,\,\LINS_a(\,(\bullet)^j\,)$,
so, by uniqueness of linearization,\\
\phx\qquad we get: \quad $\LINS_a(\,(\bullet)^j\,)\,\,=\,\,\{\,j\cdot a^{j-1}\cdot(\bullet)\,\}$.\\
Then $D_a(\,(\bullet)^j\,)\,\,=\,\,\UE(\LINS_a(\,(\bullet)^j\,)\,\,=\,\,\,j\cdot a^{j-1}\cdot(\bullet)\,$.\\
Then $\,\,((\bullet)^j)'_a\,\,=\,\,[\,D_a(\,(\bullet)^j\,)\,]\,\,=\,\,[\,j\cdot a^{j-1}\cdot(\bullet)\,]$\\
\phx\qquad\qquad\qquad $=\,\,(\,j\cdot a^{j-1}\cdot(\bullet)\,)_1\,\,=\,\,j\cdot a^{j-1}\cdot 1\,\,=\,\,j\cdot a^{j-1}$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
12-3. Show: Let $f:\R\drta\R$, $y\in\R_0^\times$.\\
\phx\qquad\qquad\qquad Assume: \quad as $x\to-\infty$, $f_x\to y$.\\
\phx\qquad\qquad\qquad Then: \qquad as $x\to-\infty$, $(1/f)_x\to1/y$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists N\in\R$ s.t., $\forall x\in\D_{1/f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,x<N\,)\,\Rightarrow(\,|(1/f)_x-(1/y)|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists N\in\R$ s.t., $\forall x\in\D_{1/f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,x<N\,)\,\Rightarrow(\,|(1/f)_x-(1/y)|<\vep\,)$.\\
Since $y\in\R_0^\times$, we get $|y|>0$.\\
Then $|y|/2>0$ and $(|y|/2)\cdot|y|>0$ and $1/|y|>0$ and $2/|y|>0$.\\
Let $\rho\,\,:=\,\,\min\,\{\,|y|/2\,,\,\vep\cdot(|y|/2)\cdot|y|\,\}$.\\
Then $\rho>0$ and $\rho\le|y|/2$ and $\rho\le\vep\cdot(|y|/2)\cdot|y|$.\\
Since $\rho\le|y|/2$, we get $-\rho\ge-|y|/2$, and so $|y|-\rho\ge|y|-(|y|/2)$.\\
Since $\rho\le\vep\cdot(|y|/2)\cdot|y|$, we get: $\rho\cdot(2/|y|)\cdot(1/|y|)\le\vep$.\\
Since \quad \quad as $x\to-\infty$, $f_x\to y$,\\
\phx\quad choose $N\in\R$ s.t., $\forall x\in\D_f$, \quad $(\,x<N\,)\,\Rightarrow(\,|f_x-y|<\rho\,)$.\\
Then $N\in\R$. \quad {\bf Want:} $\forall x\in\D_{1/f}$, $\,$ $(\,x<N\,)\,\Rightarrow(\,|(1/f)_x-(1/y)|<\vep\,)$.\\
Given $x\in\D_{1/f}$. \quad Assume: $x<N$. \quad {\bf Want:} $|(1/f)_x-(1/y)|<\vep$.\\
We have $x\in\D_{1/f}\subseteq\D_f$ and $x<N$,\\
\phx\qquad so, by choice of $N$, we get: $|f_x-y|<\rho$.\\
Since $|\bullet|$ is Lipschitz-$1$, we get: $|\,|f_x|\,-\,|y|\,|\,\,\le\,\,|f_x\,-\,y|$.\\
Then $|\,|f_x|\,-\,|y|\,|\,\,\le\,\,|f_x\,-\,y|\,\,<\,\,\rho$.\\
Then $|\,|f_x|\,-\,|y|\,|\,\,<\,\,\rho$, so $|y|-\rho\,<\,|f_x|\,<\,|y|+\rho$.\\
Then $|f_x|>|y|-\rho\ge|y|-(|y|/2)=|y|/2$.\\
Since $|f_x|>|y|/2>0$, it follows that $1/|f_x|<2/|y|$.\\
We compute: $\ds{|(1/f)_x-(1/y)|=\left|\frac{1}{f_x}-\frac{1}{y}\right|=\left|\frac{y-f_x}{f_x\cdot y}\right|=\frac{|y-f_x|}{|f_x|\cdot|y|}}$\\
Then $|(1/f)_x-(1/y)|=|y-f_x|\cdot(1/|f_x|)\cdot(1/|y|)$.\\
So, since $|y-f_x|=|f_x-y|<\rho$ and $1/|f_x|<2/|y|$ and $1/|y|>0$,\\
\phx\qquad we get: \quad $|(1/f)_x-(1/y)|<\rho\cdot(2/|y|)\cdot(1/|y|)$.\\
So, since $\rho\cdot(2/|y|)\cdot(1/|y|)\le\vep$, we get $|(1/f)_x-(1/y)|<\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
12-4. Show: Let $f:\R\drta\R$, $a\in\D_f$.\\
\phx\qquad\qquad\qquad Assume: \quad $f$ has a local strict-maximum at $a$.\\
\phx\qquad\qquad\qquad Then: \qquad $f_a^\T$ has a local strict-maximum at $0$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\exists B\in\scrb(0)$ s.t., $\forall h\in B_0^\times$, $(f_a^\T)_h<(f_a^\T)_0$.\\
Since $f$ has a local strict-maximum at $a$,\\
\phx\quad choose $C\in\scrb(a)$ s.t., $\forall x\in B_a^\times$, $f_x<f_a$.\\
Let $B:=C-a$. \quad Then $B\in\scrb(a-a)=\scrb(0)$.\\
{\bf Want:} $\forall h\in B_0^\times$, $(f_a^\T)_h<(f_a^\T)_0$.\\
Given $h\in B_0^\times$. \qquad {\bf Want:} $(f_a^\T)_h<(f_a^\T)_0$.\\
Let $x:=a+h$. \quad Then $x\in B_{a+0}^\times=B_a^\times$.\\
So, by choice of $B$, we get: $f_x<f_a$.\\
So, since $x=a+h$, we get: $f_{a+h}<f_a$. \quad Then $f_{a+h}-f_a<0$.\\
Since $a\in\D_f$, we get $f_a-f_a=0$.\\
Then $(f_a^\T)_h=f_{a+h}-f_a<0=f_a-f_a=f_{a+0}-f_a=(f_a^\T)_0$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
12-5. Show: Let $s,t\in\R^\N$.\\
\phx\qquad\qquad\qquad Assume: \quad $\I_s$ and $\I_t$ are both bounded.\\
\phx\qquad\qquad\qquad Then: \qquad $\exists$strictly-increasing $\ell\in\N^\N$ s.t.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $s\circ\ell$ and $t\circ\ell$ are both convergent.
\vskip.1in\noindent
{\it Proof:}
By properness of $\R$, since $\I_s$ is bounded, $s$ is subconvergent.\\
Choose a subsequence $\sigma$ of $s$ s.t.~$\sigma$ is convergent.\\
Since $\sigma$ is a subsequence of $s$,\\
\phx\quad choose a strictly-increasing $p\in\N^\N$ s.t.~$\sigma=s\circ p$.\\
Since $\I_{t\circ p}\subseteq\I_t$ and since $\I_t$ is bounded,\\
\phx\quad it follows that $\I_{t\circ p}$ is bounded.\\
Then, by properness of $\R$, we get: $t\circ p$ is subconvergent.\\
Choose a subsequence $\tau$ of $t\circ p$ s.t.~$\tau$ is convergent.\\
Since $\tau$ is a subsequence of $t\circ p$,\\
\phx\quad choose a strictly-increasing $q\in\N^\N$ s.t.~$\tau=t\circ p\circ q$.\\
We have $q:\N\to\N$ and $p:\N\to\N$, so $p\circ q:\N\to\N$, so $p\circ q\in\N^\N$.\\
Since $q$ and $p$ are both strictly-increasing, $p\circ q$ is strcitly-increasing.\\
Let $\ell:=p\circ q$. \quad Then $\ell\in\N^\N$ and $\ell$ is strictly-increasing.\\
{\bf Want:} $s\circ\ell$ and $t\circ\ell$ are both convergent.\\
We have $\tau=t\circ p\circ q=t\circ\ell$.\\
So, since $\tau$ is convergent, we get: $t\circ\ell$ is convergent.\\
{\bf Want:} $s\circ\ell$ is convergent.\\
Since $q\in\N^\N$ and since $q$ is strictly-increasing,\\
\phx\quad it follows that $\sigma\circ q$ is a subsequence of $\sigma$.\\
So, since $\sigma$ is convergent, by a class theorem, $\sigma\circ q$ is convergent.\\
So, since $s\circ\ell=s\circ p\circ q=\sigma\circ q$, we concude: $s\circ\ell$ is convergent. \quad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 11: Due on Tuesday 29 November}
\vskip.1in\hrule\vskip.1in\noindent
11-1. Show: Define $r:\R_0^\times\to\R$ by: $\forall x\in\R_0^\times$, $\,$ $r_x=1/x$.\\
Then: $\forall a\in\R_0^\times$, $\,$ $\ds{r'_a=-\frac{1}{a^2}}$.
\vskip.1in\noindent
{\it Proof:}
Given $a\in\R_0^\times$. \quad {\bf Want:} $\ds{r'_a=-\frac{1}{a^2}}$.\\
Let $\ds{L:=-\frac{1}{a^2}\cdot(\bullet)}$. \quad Then $\ds{[L]=L_1=-\frac{1}{a^2}}$. \quad {\bf Want:} $r'_a=[L]$.\\
{\bf Want:} $[D_ar]=[L]$. \quad {\bf Want:} $D_ar=L$.\\
{\bf Want:} $\UE(\LINS_ar)=L$. \quad {\bf Want:} $\LINS_ar=\{L\}$.\\
By uniqueness of linearization, {\bf want:} $L\in\LINS_ar$.\\
{\bf Want:} $r^\T_a-L\in\scrlito(1)$. \quad Know: $\scrbigo(2)\subseteq\scrlito(1)$. \quad {\bf Want:} $r^\T_a-L\in\scrbigo(2)$.\\
{\bf Want:} $\exists C\ge0$, $\exists\delta>0$ s.t., $\forall h\in\R$, $(|h|<\delta)\Rightarrow(|(r^\T_a-L)_h|\le C\cdot|h|^2)$.\\
Since $a\in\R_0^\times$, we get: $|a|>0$.\\
Let $\ds{C:=\frac{2}{|a|^3}}$ \quad Then $C>0$, so $C\ge0$.\\
Let $\ds{\delta:=\frac{|a|}{2}}$. \quad Then $\delta>0$.\\
{\bf Want:} $\forall h\in\R$, $(\,|h|<\delta\,)\,\Rightarrow\,(\,|(r^\T_a-L)_h|\le C\cdot|h|^2\,)$.\\
Given $h\in\R$. \quad Assume $|h|<\delta$. \quad {\bf Want:} $|(r^\T_a-L)_h|\le C\cdot|h|^2$.\\
Since $|\,|a+h|\,-\,|a|\,|\,\le\,|\,(a+h)\,-\,a\,|\,=\,|h|\,<\delta$,\\
\phx\quad we get $|a|\,-\,\delta\,<\,|a+h|\,<\,|a|\,+\,\delta$.\\
Then $\ds{|a+h|\,\ge\,|a|\,-\,\delta\,=\,|a|\,-\,\frac{|a|}{2}\,=\,\frac{|a|}{2}}$.\\
Since $\ds{|a+h|\,>\,\frac{|a|}{2}>0}$, we get: $\ds{\frac{1}{|a+h|}\,<\,\frac{1}{|a|/2}}$.\\
Then $\ds{\frac{|h|^2}{|a+h|\cdot|a|^2}\le\frac{|h|^2}{(|a|/2)\cdot|a|^2}}$.\\
We have $\ds{(r^T_a-L)_h=r_{a+h}-r_a-L_h=\frac{1}{a+h}-\frac{1}{a}-\left(-\frac{1}{a^2}\cdot h\right)}$\\
\phx\qquad\qquad\qquad $\ds{=\frac{a^2-(a+h)\cdot a+(a+h)\cdot h}{(a+h)\cdot a^2}}$\\
\phx\qquad\qquad\qquad $\ds{=\frac{a^2-a^2-h\cdot a+a\cdot h+h^2}{(a+h)\cdot a^2}\,\,=\,\,\frac{h^2}{(a+h)\cdot a^2}}$.\\
Then $\ds{|(r_a^\T-L)_h|=\frac{|h|^2}{|a+h|\cdot|a|^2}\le\frac{|h|^2}{(|a|/2)\cdot|a|^2}=\frac{2\cdot|h|^2}{|a|^3}=C\cdot|h|^2}$.\\
QED
\vskip.1in\hrule\vskip.1in\noindent
11-2. Show: Let $g:\R\drta\R$, $a\in\D'_g$.\\
Assume $g_a\ne0$. \quad Then: $\ds{\left(\frac{1}{g}\right)'_a=-\frac{g'_a}{g_a^2}}$.
\vskip.1in\noindent
{\it Proof:}
Define $r:\R_0^\times\to\R$ by: $\forall x\in\R_0^\times$, $\,$ $r_x=1/x$.\\
Then $1/g=r\circ g$. \quad {\bf Want:} $\ds{(r\circ g)'_a=-\frac{g'_a}{g_a^2}}$.\\
Since $g_a\ne0$, by HW\#11-1, we get: $\ds{r'_{g_a}=-\frac{1}{g_a^2}}$.\\
Then $\ds{r'_{g_a}\cdot g'_a=-\frac{g'_a}{g_a^2}}$. \quad
Then $\ds{(r\circ g)'_a\,\,\,\,=^*\,\,r'_{g_a}\cdot g'_a\,\,=\,\,-\frac{g'_a}{g_a^2}}$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
11-3. Show: Let $f,g:\R\drta\R$, $a\in\D'_f\cap\D'_g$.\\
Assume $g_a\ne0$. \quad Then: $\ds{\left(\frac{f}{g}\right)'_a=\frac{g_a\cdot f'_a-f_a\cdot g'_a}{g_a^2}}$.
\vskip.1in\noindent
{\it Proof:}
Since $a\in\D'_g$ and $g_a\ne0$, by HW\#11-2, we get:
$\ds{\left(\frac{1}{g}\right)'_a=-\frac{g'_a}{g_a^2}}$.\\
Let $\ds{h:=\frac{1}{g}}$. \quad Then $\ds{h'_a=-\frac{g'_a}{g_a^2}}$. \quad Also, $\ds{h_a=\frac{1}{g_a}}$.\\
Then $\,\,\,\ds{\left(\frac{f}{g}\right)'_a\,\,=\,\,(f\cdot h)'_a\,\,\,\,=^*\,\,f'_a\cdot h_a+f_a\cdot h'_a}$\\
\phx\qquad\qquad\qquad $\ds{=\,\,\frac{f'_a}{g_a}-\frac{f_a\cdot g'_a}{g_a^2}\,\,=\,\,\frac{g_a\cdot f'_a-f_a\cdot g'_a}{g_a^2}.}$ \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
11-4. Show: Let $U,V\subseteq\R$ both be open. Then:\\
\phx\quad (i) $U\cup V$ is open \qquad\qquad and \qquad\qquad (ii) $U\cap V$ is open.
\vskip.1in\noindent
{\it Proof:}
{\it Proof of (i):}\\
{\bf Want:} $\forall a\in U\cup V$, $\exists B\in\scrb(a)$ s.t.~$B\subseteq U\cup V$.\\
Given $a\in U\cup V$. \quad {\bf Want:} $\exists B\in\scrb(a)$ s.t.~$B\subseteq U\cup V$.\\
Since $a\in U\cup V$, choose $T\in\{U,V\}$ s.t.~$a\in T$.\\
Since $T\in\{U,V\}$, it follows that $T$ is open and that $T\subseteq U\cup V$.\\
Since $T$ is open and $a\in T$, choose $B\in\scrb(a)$ s.t.~$B\subseteq T$.\\
Then $B\in\scrb(a)$. \quad {\bf Want:} $B\subseteq U\cup V$. \quad
We have $B\subseteq T\subseteq U\cup V$.\\
{\it End of proof of (i).}\\
\\
{\it Proof of (ii):}\\
{\bf Want:} $\forall a\in U\cap V$, $\exists B\in\scrb(a)$ s.t.~$B\subseteq U\cap V$.\\
Given $a\in U\cap V$. \quad {\bf Want:} $\exists B\in\scrb(a)$ s.t.~$B\subseteq U\cap V$.\\
Since $a\in U\cap V\subseteq U$ and since $U$ is open,\\
\phx\qquad choose $\lambda>0$ s.t.~$B(a,\lambda)\subseteq U$.\\
Since $a\in U\cap V\subseteq V$ and since $V$ is open,\\
\phx\qquad choose $\mu>0$ s.t.~$B(a,\mu)\subseteq V$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \quad Then $\delta>0$ and $\delta\le\lambda$ and $\delta\le\mu$.\\
Let $B:=B(a,\delta)$. \quad Then $B\in\scrb(a)$. \quad {\bf Want:} $B\subseteq U\cap V$.\\
Since $\delta\le\lambda$, we get $B(a,\delta)\subseteq B(a,\lambda)$.\\
Since $\delta\le\mu$, we get $B(a,\delta)\subseteq B(a,\mu)$.\\
Since $B=B(a,\delta)\subseteq B(a,\lambda)\subseteq U$ and $B=B(a,\delta)\subseteq B(a,\mu)\subseteq V$,\\
\phx\qquad we get: $B\subseteq U\cap V$.\\
{\it End of proof of (ii).} \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
11-5. Show: Let $f:\R\drta\R$. \quad
Assume: $\forall$open $U\subseteq\R$, $\,$ $f^*U$ is open.\\
Then: $f$ is continuous.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall a\in\D_f$, $f$ is continuous at $a$.\\
Given $a\in\D_f$. \quad {\bf Want:} $f$ is continuous at $a$.\\
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Let $U:=B(f_a,\vep)$. \quad By a class theorem, $U$ is open.\\
Then, by assumption, $f^*U$ is open.\\
Since $|f_a-f_a|=0<\vep$, we get: $f_a\in B(f_a,\vep)$.\\
Since $f_a\in B(f_a,\vep)=U$, we get: $a\in f^*U$.\\
Since $f^*U$ is open and $a\in f^*U$, choose $B\in\scrb(a)$ s.t.~$B\subseteq f^*U$.\\
Since $B\in\scrb(a)$, choose $\delta>0$ s.t.~$B=B(a,\delta)$. \quad Then $\delta>0$.\\
{\bf Want:} $\forall x\in\D_f$, $\,$ $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Given $x\in\D_f$. \quad Assume: $|x-a|<\delta$. \quad {\bf Want:} $|f_x-f_a|<\vep$.\\
Since $|x-a|<\delta$, we get: $x\in B(a,\delta)$.\\
Since $x\in B(a,\delta)=B\subseteq f^*U$, we get: $f_x\in U$.\\
Since $f_x\in U=B(f_a,\vep)$, we get: $|f_x-f_a|<\vep$. \quad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 10: Due on Tuesday 22 November}
\vskip.1in\hrule\vskip.1in\noindent
10-1. Show: Let $w,x\in\R^\N$, $\ell\in\N^\N$, $q\in\R$.\\
Assume: $($ $\ell$ is strictly-increasing $)$ $\&$ $(\,w\circ\ell\to q\,)$.\\
Assume: $\forall j\in\N$, $|w_j-x_j|<1/j$.\\
Then: $x\circ\ell\to q$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(x\circ\ell)_j-q|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(x\circ\ell)_j-q|<\vep\,)$.\\
Since $w\circ\ell\to q$, choose $L\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(w\circ\ell)_j-q|<\vep/2\,)$.\\
By the AP, choose $M\in\N$ s.t.~$M>2/\vep$.\\
Let $K:=\max\{L,M\}$. \qquad Then $K\ge L$ and $K\ge M$ and $K\in\N$.\\
{\bf Want:} $\forall j\in\N$, \quad $(\,j\ge K\,)\,\Rightarrow\,(\,|(x\circ\ell)_j-q|<\vep\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $|(x\circ\ell)_j-q|<\vep$.\\
Since $j\ge K\ge L$, by choice of $L$, we get: $|(w\circ\ell)_j-q|<\vep/2$.\\
By assumption, $|x_{\ell_j}-w_{\ell_j}|<1/\ell_j$.\\
By a class theorem, $\ell_j\ge j$. \quad Then $\ell_j\ge j\ge K\ge M>2/\vep$.\\
Since $\ell_j\ge2/\vep>0$, we get: $1/\ell_j<\vep/2$.\\
Then $|x_{\ell_j}-w_{\ell_j}|<1/\ell_j<\vep/2$, so $|x_{\ell_j}-w_{\ell_j}|<\vep/2$.\\
Then $\,\,|(x\circ\ell)_j-q|\le|(x\circ\ell)_j-(w\circ\ell)_j|+|(w\circ\ell)_j-q|$\\
\phx\qquad\qquad\qquad\qquad $=|x_{\ell_j}-w_{\ell_j}|+|(w\circ\ell)_j-q|$\\
\phx\qquad\qquad\qquad\qquad $<(\vep/2)+(\vep/2)\quad=\quad\vep$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
10-2. Show: Let $f,g:\R\drta\R$.\\
Assume: as $x\to\infty$, $f_x\to-\infty$.\\
Assume: as $x\to\infty$, $g_x\to-\infty$.\\
Then: as $x\to\infty$, $(f\cdot g)_x\to\infty$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall M\in\R$, $\exists L\in\R$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,x>L\,)\,\Rightarrow\,(\,(f\cdot g)_x>M\,)$.\\
Given $M\in\R$. \quad {\bf Want:} $\exists L\in\R$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,x>L\,)\,\Rightarrow\,(\,(f\cdot g)_x>M\,)$.\\
Let $N:=-\sqrt{\max\{M,0\}}$. \quad Then $N\in\R$ and $N\le0$.\\
Also, $N^2=\max\{M,0\}$, so $N^2\ge M$.\\
Since \quad as $x\to\infty$, $f_x\to-\infty$, \quad choose $A\in\R$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,x>A\,)\,\Rightarrow\,(\,f_x<N\,)$.\\
Since \quad as $x\to\infty$, $g_x\to-\infty$, \quad choose $B\in\R$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,x>B\,)\,\Rightarrow\,(\,g_x<N\,)$.\\
Let $L:=\max\{A,B\}$. \quad Then $L\ge A$ and $L\ge B$ and $L\in\R$.\\
{\bf Want:} $\forall x\in\R$, \quad $(\,x>L\,)\,\Rightarrow\,(\,(f\cdot g)_x>M\,)$.\\
Given $x\in\R$. \quad Assume $x>L$. \quad {\bf Want:} $(f\cdot g)_x>M$.\\
Since $x>L\ge A$, by choice of $A$, we have: $f_x<N$.\\
Then $f_x<N\le0$, so $-f_x>-N\ge0$.\\
Since $x>L\ge B$, by choice of $B$, we have: $g_x<N$.\\
Then $g_x<N\le0$, so $-g_x>-N\ge0$.\\
Since $-f_x>-N\ge 0$ and $-g_x>-N\ge0$,\\
\phx\quad we get $(-f_x)\cdot(-g_x)>(-N)\cdot(-N)$, so $f_x\cdot g_x>N^2$.\\
Then $(f\cdot g)_x=f_x\cdot g_x>N^2\ge M$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
10-3. Show: Let $f:=(\bullet)^3$. Then $D_2f=3\cdot2^2\cdot(\bullet)$.
\vskip.1in\noindent
{\it Proof:}
Let $x:=2$. \quad {\bf Want:} $D_xf=3\cdot x^2\cdot(\bullet)$.\\
Since $\D_f=\R$, we see that $\D_{f_x^\T}=\R$.\\
We have: $\forall h\in\R$,\\
\phx\qquad $(f_x^\T-3\cdot x^2\cdot(\bullet))_h=f_{x+h}-f_x-3x^2h$\\
\phx\qquad\qquad\qquad\qquad\qquad $=(x+h)^3-x^3-3x^2h$\\
\phx\qquad\qquad\qquad\qquad\qquad $=x^3+3x^2h+3xh^2+h^3-x^3-3x^2h$\\
\phx\qquad\qquad\qquad\qquad\qquad $=3xh^2+h^3\,\,=\,\,((3\cdot x\cdot(\bullet)^2+(\bullet)^3)_h$,\\
and so $f_x^\T-3\cdot x^2\cdot(\bullet)=3\cdot x\cdot(\bullet)^2+(\bullet)^3$.\\
We have $3\cdot x\cdot(\bullet)^2\in\scrh_2\subseteq\scrbigo(2)\subseteq\scrlito(1)$.\\
Also, $(\bullet)^3\in\scrh_3\subseteq\scrbigo(3)\subseteq\scrlito(1)$.\\
Since $3\cdot x\cdot(\bullet)^2\in\scrlito(1)$ and $(\bullet)^3\in\scrlito(1)$,\\
\phx\quad and since $\scrlito(1)$ is linearly closed, we get:\\
\phx\qquad\qquad $3\cdot x\cdot(\bullet)^2+(\bullet)^3\in\scrlito(1)$.\\
Since $f_x^\T-3\cdot x^2\cdot(\bullet)=3\cdot x\cdot(\bullet)^2+(\bullet)^3\in\scrlito(1)$,\\
\phx\quad we get $3\cdot x^2\cdot(\bullet)\in\LINS_xf$.\\
Then, by uniqueness of linearization, $\LINS_xf=\{3\cdot x^2\cdot(\bullet)\}$.\\
Then $D_xf\,=\,\UE(\LINS_xf)\,=\,\UE\{3\cdot x^2\cdot(\bullet)\}\,\,=\,\,3\cdot x^2\cdot(\bullet)$.
QED
\vskip.1in\hrule\vskip.1in\noindent
10-4. Show: Let $f:=(\bullet)^3$. Then $\forall x\in\R$, $f'_x=3x^2$.
\vskip.1in\noindent
{\it Proof:}
Since $\D_f=\R$, we see that $\D_{f_x^\T}=\R$.\\
We have: $\forall h\in\R$,\\
\phx\qquad $(f_x^\T-3\cdot x^2\cdot(\bullet))_h=f_{x+h}-f_x-3x^2h$\\
\phx\qquad\qquad\qquad\qquad\qquad $=(x+h)^3-x^3-3x^2h$\\
\phx\qquad\qquad\qquad\qquad\qquad $=x^3+3x^2h+3xh^2+h^3-x^3-3x^2h$\\
\phx\qquad\qquad\qquad\qquad\qquad $=3xh^2+h^3\,\,=\,\,((3\cdot x\cdot(\bullet)^2+(\bullet)^3)_h$,\\
and so $f_x^\T-3\cdot x^2\cdot(\bullet)=3\cdot x\cdot(\bullet)^2+(\bullet)^3$.\\
We have $3\cdot x\cdot(\bullet)^2\in\scrh_2\subseteq\scrbigo(2)\subseteq\scrlito(1)$.\\
Also, $(\bullet)^3\in\scrh_3\subseteq\scrbigo(3)\subseteq\scrlito(1)$.\\
Since $3\cdot x\cdot(\bullet)^2\in\scrlito(1)$ and $(\bullet)^3\in\scrlito(1)$,\\
\phx\quad and since $\scrlito(1)$ is linearly closed, we get:\\
\phx\qquad\qquad $3\cdot x\cdot(\bullet)^2+(\bullet)^3\in\scrlito(1)$.\\
Since $f_x^\T-3\cdot x^2\cdot(\bullet)=3\cdot x\cdot(\bullet)^2+(\bullet)^3\in\scrlito(1)$,\\
\phx\quad we get $3\cdot x^2\cdot(\bullet)\in\LINS_xf$.\\
Then, by uniqueness of linearization, $\LINS_xf=\{3\cdot x^2\cdot(\bullet)\}$.\\
Then $D_xf\,=\,\UE(\LINS_xf)\,=\,\UE\{3\cdot x^2\cdot(\bullet)\}\,\,=\,\,3\cdot x^2\cdot(\bullet)$.\\
Then $f'_x\,=\,[D_xf]\,=\,[3\cdot x^2\cdot(\bullet)]$\\
\phx\qquad\qquad $=\!\!\!\!\!{}^*\,\,\,\,\,\,(3\cdot x^2\cdot(\bullet))_1\,\,=\,\,3\cdot x^2\cdot 1\,\,=\,\,3x^2$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
10-5. Show: Let $f:\R\drta\R$, $a,c\in\R$.\\
Then $D_a(c\cdot f)\,\,\,\,=^*\,\,c\cdot(D_af)$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $($ $c\cdot(D_af)\ne\frownie$ $)$ $\Rightarrow$ $($ $D_a(c\cdot f)=c\cdot(D_af)$ $)$.\\
Assume: $c\cdot(D_af)\ne\frownie$. \quad {\bf Want:} $D_a(c\cdot f)=c\cdot(D_af)$.\\
Since $c\cdot(D_af)\frownie$, we get: $D_af\ne\frownie$. \quad Then $D_af\in\LINS_af$.\\
Let $L:=D_af$. \quad Then $L\in\LINS_af$, so $f^\T_a-L\in\scrlito(1)$.\\
So, since $\scrlito(1)$ is linearly closed, we get: $c\cdot(f^\T_a-L)\in\scrlito(1)$.\\
Then $(c\cdot f)_a^\T-c\cdot L=c\cdot f_a^\T-c\cdot L=c\cdot(f^\T_a-L)\in\scrlito(1)$,\\
\phx\quad and so $c\cdot L\in\LINS_a(c\cdot f)$.\\
Then, by uniqueness of linearization, we get $\LINS_a(c\cdot f)=\{c\cdot L\}$.\\
Then $D_a(c\cdot f)\,=\,\UE(\LINS_a(c\cdot f))\,=\,\UE\{c\cdot L\}\,\,=\,\,c\cdot L$.\\
Then $D_a(c\cdot f)=c\cdot L=c\cdot(D_af)$. \qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 9: Due on Tuesday 15 November}
\vskip.1in\hrule\vskip.1in\noindent
9-1. Show: Define $f:\R\drta\R$ by: \quad $\forall x\in\R$, $\,$ $f_x\,=\,1/x$.\\
Then: \quad as $x\to-\infty$, $\,$ $f_x\to0$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists N\in\R$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,x<N\,)\,\Rightarrow\,(\,|f_x-0|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists N\in\R$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,x<N\,)\,\Rightarrow\,(\,|f_x-0|<\vep\,)$.\\
Let $N:=-1/\vep$. \qquad Then $-1/N=\vep$.\\
Since $\vep>0$, we get: $N\in\R$ and $N<0$.\\
{\bf Want:} $\forall x\in\D_f$, \quad $(\,x<N\,)\,\Rightarrow\,(\,|f_x-0|<\vep\,)$.\\
Given $x\in\D_f$. \quad Assume: $x<N$. \quad {\bf Want:} $|f_x-0|<\vep$.\\
We have $x<N<0$, so $x<0$, so $1/x<0$, so $|1/x|=-1/x$.\\
Then $|f_x-0|=|f_x|=|1/x|=-1/x$. \quad Also, $\vep=-1/N$.\\
{\bf Want:} $-1/x<-1/N$.\\
We have $x<N<0$, so $-x>-N>0$, so $1/(-x)<1/(-N)$.\\
Then $-1/x=1/(-x)<1/(-N)=-1/N$.
QED
\vskip.1in\hrule\vskip.1in\noindent
9-2. Show: Let $D\subseteq\R$, $s\in(\R^D)^\N$, $f\in\R^D$.\\
Assume: \quad $\forall j\in\N$, $s_j$ is continuous.\\
Assume: \quad $s\to f$ uniformly.\\
Then: \qquad $f$ is continuous.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall a\in\D_f$, $f$ is continuous at $a$.\\
Given $a\in\D_f$. \quad {\bf Want:} $f$ is continuous at $a$.\\
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|[f(x)]-[f(a)]|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|[f(x)]-[f(a)]|<\vep\,)$.\\
Since $s\to f$ uniformly, choose $K\in\N$ s.t., $\forall j\in\N$, $\forall x\in D$\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|[s_j(x)]-[f(x)]|<\vep/3\,)$.\\
By assumption, $s\in(\R^D)^\N$, so $s_K\in\R^D$, so $\D_{s_K}=D$.\\
Also, $f\in\R^D$, so $\D_f=D$.\\
By assumption, $\forall j\in\N$, $s_j$ is continuous. \quad Then $s_K$ is continuous.\\
So, since $a\in\D_f=D=\D_{s_K}$, we see that $s_K$ is continuous at $a$.\\
Choose $\delta>0$ s.t., $\forall x\in\D_{s_K}$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|[s_K(x)]-[s_K(a)]|<\vep/3\,)$.\\
Then $\delta>0$. \quad {\bf Want:} $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|[f(x)]-[f(a)]|<\vep\,)$.\\
Given $x\in\D_f$. \quad Assume: $|x-a|<\delta$. \quad {\bf Want:} $|[f(x)]-[f(a)]|<\vep$.\\
We have $x\in D_f=D$, so $x\in D$. \quad Then $x\in D=\D_{s_K}$, so $x\in\D_{s_K}$.\\
Since $K\ge K$ and $x\in D$, by choice of $K$, $|[s_K(x)]-[f(x)]|<\vep/3$.\\
Since $|x-a|<\delta$, by choice of $\delta$, $|[s_K(x)]-[s_K(a)]|<\vep/3$.\\
Since $K\ge K$ and $a\in D$, by choice of $K$, $|[s_K(a)]-[f(a)]|<\vep/3$.\\
Then $|[f(x)]-[f(a)]|$\\
\phx\qquad\quad $\le|[f(x)]-[s_K(x)]|+|[s_K(x)]-[s_K(a)]||[s_K(a)]-[f(a)]|$\\
\phx\qquad\quad $=|[s_K(x)]-[f(x)]|+|[s_K(x)]-[s_K(a)]||[s_K(a)]-[f(a)]|$\\
\phx\qquad\quad $<(\vep/3)+(\vep/3)+(\vep/3)=\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
9-3. Show: Let $f,g:\R\drta\R$, $k\in\N_0$.\\
Assume: \quad $(\,\,f\in\scrlito(k)\,\,)\,\,\&\,\,(\,\,\hbox{near }0,\,\,f=g\,\,)$.\\
Then: \qquad\quad $g\in\scrlito(k)$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\delta\,)\,\Rightarrow\,(\,|g_x|<\vep\cdot|x|^k\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\delta\,)\,\Rightarrow\,(\,|g_x|<\vep\cdot|x|^k\,)$.\\
Since $f\in\scrlito(k)$, choose $\alpha>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\alpha\,)\,\Rightarrow\,(\,|f_x|<\vep\cdot|x|^k\,)$.\\
Since \quad near $0$, $f=g$, \quad choose $\beta>0$ s.t.,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad on $B(0,\delta)$, \quad $f=g$.\\
Then $\forall x\in B(0,\delta)$, we have $f_x=g_x$.\\
Let $\delta:=\min\{\alpha,\beta\}$. \quad Then $\delta>0$.\\
{\bf Want:} $\forall x\in\R$, $(\,|x|<\delta\,)\,\Rightarrow\,(\,|g_x|<\vep\cdot|x|^k\,)$.\\
Given $x\in\R$. \quad Assume: $|x|<\delta$. \quad {\bf Want:} $|g_x|<\vep\cdot|x|^k$.\\
Since $|x|<\delta\le\alpha$, by choice of $\alpha$, we get: $|f_x|<\vep\cdot|x|^k$.\\
Since $|x-0|<\delta\le\beta$, we get: $x\in B(0,\beta)$, and so $f_x=g_x$.\\
Then $|g_x|=|f_x|<\vep\cdot|x|^k$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
9-4. Show: Let $s,t\in\R^\N$.\\
Assume: \quad $(\,\,s\to\infty\,\,)\,\,\&\,\,(\,\,\forall j\in\N,\,\,t_j\ge s_j\,\,)$.\\
Then: \qquad\quad $t\to\infty$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall M\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,t_j>M\,)$.\\
Given $M\in\R$. \quad {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,t_j>M\,)$.\\
Since $s\to\infty$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>M\,)$.\\
Then $K\in\N$.\\
{\bf Want:} $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,t_j>M\,)$.\\
Given $j\in\N$. \quad Assume: $j\ge K$. \quad {\bf Want:} $t_j>M$.\\
Since $j\in\N$, by assumption, we get: $t_j\ge s_j$.\\
Since $j\ge K$, by choice of $K$, we get: $s_j>M$.\\
Then $t_j\ge s_j>M$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
9-5. Show:\\
Let $a\,:=\,(\,1\,,\,1/2\,,\,1/2\,,\,1/4\,,\,1/4\,,\,1/4\,,\,1/4\,,\,1/8\,,\,\ldots)$.\\
\phx NOTE: The terms of $a$ are, in order:\\
\phx\qquad\qquad\qquad a single ``$1$'', \quad then two ``$1/2$''s,\\
\phx\qquad\qquad\qquad then four ``$1/4$''s, \quad then eight ``$1/8$''s,\\
\phx\qquad\qquad\qquad then sixteen ``$1/16$''s, \quad then thirty-two ``$1/32$''s, \quad {\it etc.}\\
Let $b\,:=\,(\,1\,,\,1/2\,,\,1/3\,,\,1/4\,,\,1/5\,,\,1/6\,,\,1/7\,,\,1/8\,,\,\ldots)$.\\
Define $s,t\in\R^\N$ by: $\forall j\in\N$,\\
\phx\qquad $s_j=a_1+\cdots+a_j$ \quad and \quad $t_j=b_1+\cdots+b_j$.\\
Then: $\,\,\,$(i) $\I_s$ is unbounded \qquad\quad and\\
\phx\qquad $\,\,\,\,\,$(ii) $\forall j\in\N$, $\,$ $t_j>s_j/2$ \qquad and\\
\phx\qquad $\,\,\,\,$(iii) $t\to\infty$.
\vskip.1in\noindent
{\it Proof:}
{\it Proof of (i):}\\
Define $m\in\N^\N$ by: $\forall j\in\N$, $m_j=1+2+4+8+16+\cdots+2^{j-1}$.\\
By a class theorem, $\forall j\in\N$, $s_{m_j}=j$.\\
Then: $\forall j\in\N$, $j=s_{m_j}\in\I_s$.\\
Since, $\forall j\in\N$, $j\in\I_s$, it follows that $\N\subseteq\I_s$.\\
Assume $\I_s$ is bounded. \quad {\bf Want:} Contradiction.\\
Choose $B\in\scrb_\R$ s.t.~$\I_s\subseteq B$.\\
Choose $a\in\R$ and $r>0$ s.t.~$B=B(a,r)$.\\
By the AP, choose $j\in\N$ s.t.~$j>a+r$.\\
Since $j\in\N\subseteq\I_s\subseteq B=B(a,r)=(a-r;a+r)<a+r<j$,\\
\phx\qquad we conclude that $j<j$. \quad Contradiction.\\
{\it End of proof of (i).}\\
\\
{\it Proof of (ii):}\\
Given $j\in\N$. \quad {\bf Want:} $t_j>s_j/2$.\\
By a class theorem, $\forall i\in\N$, $b_i>s_i/2$.\\
Then all of the following are true:\\
\phx\quad $b_1>a_1/2$, \quad $b_2>a_2/2$, \quad $b_3>a_3/2$, \quad $\cdots$ \quad $b_j>a_j/2$.\\
Then $b_1+\cdots+b_j>(a_1/2)+\cdots+(a_j/2)$.\\
Then $t_j=b_1+\cdots+b_j>(a_1/2)+\cdots+(a_j/2)=(a_1+\cdots+a_j)/2=s_j/2$. \quad
{\it End of proof of (ii).}\\
\\
{\it Proof of (iii):}\\
Since $\forall j\in\N$, $a_j>0$, it follows that $s$ is strictly-increasing.\\
Then $s$ is semi-increasing. \qquad
By (i), $\I_s$ is unbounded.\\
Then, by a class theorem, $s\to\infty$.\\
By (ii), we have: $\forall j\in\N$, $t_j>(s/2)_j$.\\
So, by HW\#9-4, it suffices to show: $s/2\to\infty$.\\
{\bf Want:} $\forall M\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,(s/2)_j>M\,)$.\\
Given $M\in\R$. \quad {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,(s/2)_j>M\,)$.\\
Since $s\to\infty$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>2\cdot M\,)$.\\
Then $K\in\N$. \qquad
{\bf Want:} $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,(s/2)_j>M\,)$.\\
Given $j\in\N$. \quad Assume: $j\ge K$. \quad {\bf Want:} $(s/2)_j>M$.\\
Since $j\ge K$, by choice of $K$, we get: $s_j>2\cdot M$.\\
Then $s_j/2>(2\cdot M)/2$.\\
Then $(s/2)_j=s_j/2>(2\cdot M)/2=M$.\\
{\it End of proof of (iii).} \quad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 8: Due on Wednesday 9 November}
\vskip.1in\hrule\vskip.1in\noindent
8-1. Show: Let $f\in\scrlito(3)$, $g\in\scrlito(4)$. \quad
Then: $f\cdot g\in\scrlito(7)$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\delta\,)\,\Rightarrow\,(\,|(f\cdot g)_x|\le\vep\cdot|x|^7\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\delta\,)\,\Rightarrow\,(\,|(f\cdot g)_x|\le\vep\cdot|x|^7\,)$.\\
Since $f\in\scrlito(3)$, choose $\alpha>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\alpha\,)\,\Rightarrow\,(\,|f_x|\le\sqrt{\vep}\cdot|x|^3\,)$.\\
Since $g\in\scrlito(4)$, choose $\beta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x|<\beta\,)\,\Rightarrow\,(\,|g_x|\le\sqrt{\vep}\cdot|x|^4\,)$.\\
Let $\delta:=\min\{\alpha,\beta\}$. \quad Then $\delta>0$ and $\delta\le\alpha$ and $\delta\le\beta$.\\
{\bf Want:} $\forall x\in\R$, $\,$ $(\,|x|<\delta\,)\,\Rightarrow\,(\,|(f\cdot g)_x|\le\vep\cdot|x|^7\,)$.\\
Given $x\in\R$. \quad Assume $|x|<\delta$. \quad {\bf Want:} $|(f\cdot g)_x|\le\vep\cdot|x|^7$.\\
Since $|x|<\delta\le\alpha$, by choice of $\alpha$, we get: $|f_x|\le\sqrt{\vep}\cdot|x|^3$.\\
Since $|x|<\delta\le\beta$, by choice of $\beta$, we get: $|g_x|\le\sqrt{\vep}\cdot|x|^4$.\\
Since $0\le|f_x|\le\sqrt{\vep}\cdot|x|^3$ and $0\le|g_x|\le\sqrt{\vep}\cdot|x|^4$,\\
\phx\qquad we get: \quad $\,\,\,|f_x|\cdot|g_x|\le\sqrt{\vep}\cdot|x|^3\cdot\sqrt{\vep}\cdot|x|^4$.\\
Then $|(f\cdot g)_x|=|f_x\cdot g_x|$\\
\phx\qquad\qquad\qquad $=|f_x|\cdot|g_x|\le\sqrt{\vep}\cdot|x|^3\cdot\sqrt{\vep}\cdot|x|^4=\vep\cdot|x|^7$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
8-2. Show: Let $s\in\R^\N$.\\
Assume $\I_s$ is bounded. \quad Then $s$ is subconvergent.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\exists$subsequence $t$ of $s$ s.t.~$t$ is convergent.\\
By a class theorem, choose a subsequence $t$ of $s$ s.t.~$t$ is semi-monotone.\\
Then $t$ is a subsequence of $s$. \quad {\bf Want:} $t$ is convergent.\\
Since $t$ is a subsequence of $s$, we get: $\I_t\subseteq\I_s$.\\
So, since $\I_s$ is bounded, we get: $\I_t$ is bounded.\\
Since $t$ is semi-monontone and $\I_t$ is bounded,\\
\phx\quad by a class theorem, we get: $t$ is convergent. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
8-3. Show: Let $s\in\R^\N$.\\
Assume $s$ is convergent. \quad Then $s$ is Cauchy.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,i,j\ge K\,)\,\Rightarrow\,(\,|s_i-s_j|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,i,j\ge K\,)\,\Rightarrow\,(\,|s_i-s_j|<\vep\,)$.\\
Since $s$ is convergent, choose $z\in\R$ s.t.~$s\to z$.\\
Since $s\to z$, choose $K\in\N$ s.t., $\forall k\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,k\ge K\,)\,\Rightarrow\,(\,|s_k-z|<\vep/2\,)$.\\
Then $K\in\N$. \quad {\bf Want:} $\forall i,j\in\N$, $(\,i,j\ge K\,)\,\Rightarrow\,(\,|s_i-s_j|<\vep\,)$.\\
Given $i,j\in\N$. \quad Assume $i,j\ge K$. \quad {\bf Want:} $|s_i-s_j|<\vep$.\\
Since $i\ge K$, by choice of $K$, we get: $|s_i-z|<\vep/2$.\\
Since $j\ge K$, by choice of $K$, we get: $|s_j-z|<\vep/2$, so $|z-s_j|<\vep/2$.\\
Then $|s_i-s_j|\le|s_i-z|+|z-s_j|<(\vep/2)+(\vep/2)=\vep$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
8-4. Show: Let $s\in\R^\N$.\\
Assume $s$ is Cauchy and subconvergent. \quad Then $s$ is convergent.
\vskip.1in\noindent
{\it Proof:}
Since $s$ is subconvergent,\\
\phx\qquad\qquad\qquad choose a subsequence $t$ of $s$ s.t.~$t$ is convergent.\\
Since $t$ is convergent, choose $z\in\R$ s.t.~$t\to z$. \quad
{\bf Want:} $s\to z$.\\
{\bf Want:} $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|s_j-z|<\vep\,)$.\\
Given $\vep>0$. \quad {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|s_j-z|<\vep\,)$.\\
Since $s$ is Cauchy, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,i,j\ge K\,)\,\Rightarrow\,(\,|s_i-s_j|<\vep/2\,)$.\\
Then $K\in\N$. \quad {\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,|s_j-z|<\vep\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $|s_j-z|<\vep$.\\
Since $t\to z$, choose $L\in\N$ s.t., $\forall i\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,i\ge L\,)\,\Rightarrow\,(\,|t_i-z|<\vep/2\,)$.\\
Let $i:=\max\{K,L\}$. \quad Then $i\in\N$ and $i\ge K$ and $i\ge L$.\\
Since $t$ is a subsequence of $s$,\\
\phx\qquad choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
Since $\ell\in\N^\N$, and $i\in\N$, we get: $\ell_i\in\N$.\\
By a homework problem, we get: \quad $\ell_i\ge i$.\\
Let $m:=\ell_i$. \qquad Then \quad $m\in\N$ \quad and \quad $m\ge i$.\\
Since $m\ge i\ge K$, we get: $m\ge K$.\\
Since $s_m=s_{\ell_i}=(s\circ\ell)_i=t_i$, we get $s_m=t_i$.\\
Since $i\ge L$, by choice of $L$, we get: $|t_i-z|<\vep/2$.\\
Since $m,j\ge K$, by choice of $K$, we get: $|s_m-s_j|<\vep/2$.\\
Then $|s_j-t_i|=|s_j-s_m|=|s_m-s_j|<\vep/2$, \quad so $|s_j-t_i|<\vep/2$.\\
Then $|s_j-z|\le|s_j-t_i|+|t_i-z|<(\vep/2)+(\vep/2)=\vep$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
8-5. Show: Let $s\in\R^\N$.\\
Assume $s$ is Cauchy. \quad Then $s$ is convergent.
\vskip.1in\noindent
{\it Proof:}
By a class theorem, \quad since $s$ is Cauchy, \quad $\I_s$ is bounded.\\
Then, by HW\#8-2, \quad we get: $\,$ $s$ is subconvergent.\\
Since $s$ is Cauchy and subconvergent,\\
\phx\qquad by HW\#8-4, \quad we get: $\,$ $s$ is convergent. \quad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 7: Due on Tuesday 1 November}
\vskip.1in\hrule\vskip.1in\noindent
7-1. Show: Let $k\in\N^\N$. \quad Assume $k$ is strictly-increasing.\\
Then: \quad $\forall j\in\N$, $\,$ $k_j\ge j$.
\vskip.1in\noindent
{\it Proof:}
Let $S:=\{j\in\N\,|\,k_j\ge j\}$. \quad {\bf Want:} $S=\N$.\\
Since $k_1\in\N\ge1$, we get $1\in S$.\\
By the PMI, it suffice to show: \quad $\forall k\in S$, $\,$ $k+1\in S$.\\
Given $k\in S$. \quad {\bf Want:} $k+1\in S$.\\
Know: $k_j\ge j$. \quad {\bf Want:} $k_{j+1}\ge j+1$.\\
Since $k$ is strictly increasing, we have: $k_{j+1}>k_j$.\\
So, since $k_j,k_{j+1}\in\Z$, we get: $k_{j+1}\ge k_j+1$.\\
Since  $k_j\ge j$, by adding $1$ to both sides, we get: $k_j+1\ge j+1$.\\
Then $k_{j+1}\ge k_j+1\ge j+1$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
7-2. Show: Let $s,t\in\R$. \quad Assume $s<t$.\\
Then: \quad $\exists x\in\Q$ s.t.~$s<x<t$.
\vskip.1in\noindent
{\it Proof:}
We have $t-s>0$. \quad
By the AP, choose $j\in\N$ s.t.~$j>1/(t-s)$.\\
Then $j\cdot(t-s)>1$, so $j\cdot t-j\cdot s>1$.\\
Let $a:=j\cdot s$ and $b:=j\cdot t$. \quad Then $b-a>1$.\\
Then, by a class theorem, choose $k\in\Z$ s.t.~$a<k<b$.\\
Since $k\in\Z$ and $j\in\N$, we get $k/j\in\Q$.\\
Let $x:=k/j$. \quad Then $x\in\Q$. \quad {\bf Want:} $s<x<t$.\\
We have $j\cdot x=k$.\\
Since $a<k<b$ and since $j\in\N>0$, we get $a/j<k/j<b/j$.\\
Then $(j\cdot s)/j<(j\cdot x)/j<(j\cdot t)/j$. \quad Then $s<x<t$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
7-3. Show: Let $s,t,u\in\R^\N$, $a\in\R$.\\
Assume: $\forall j\in\N$, $s_j\le t_j\le u_j$.\\
Assume: $s\to a$ and $u\to a$.\\
Then: $t\to a$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,|t_j-a|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,|t_j-a|<\vep\,)$.\\
Since $s\to a$, choose $L\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge L\,)\,\Rightarrow\,(\,|s_j-a|<\vep\,)$.\\
Since $u\to a$, choose $M\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge M\,)\,\Rightarrow\,(\,|u_j-a|<\vep\,)$.\\
Let $K:=\max\{L,M\}$. \quad Then $K\ge L$ and $K\ge M$ and $K\in\N$.\\
{\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,|t_j-a|<\vep\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $|t_j-a|<\vep$.\\
Since $j\ge K\ge L$, by choice of $L$, we get $|s_j-a|<\vep$.\\
Then $a-\vep<s_j<a+\vep$.\\
Since $j\ge K\ge M$, by choice of $M$, we get $|u_j-a|<\vep$.\\
Then $a-\vep<u_j<a+\vep$.\\
By assumption, $s_j\le t_j\le u_j$.\\
We have $a-\vep<s_j\le t_j$, \quad so $a-\vep<t_j$.\\
We have $t_j\le u_j<a+\vep$, \quad so $t_j<a+\vep$.\\
Then $a-\vep<t_j<a+\vep$, \quad so $|t_j-a|<\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
7-4. Show: Let $f,g:\R\drta\R$, $a\in\R$.\\
Assume: as $x\to a$, $f_x\to-\infty$.\\
Assume: as $x\to a$, $g_x\to-\infty$.\\
Then: as $x\to a$, $(f\cdot g)_x\to\infty$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall M\in\R$, $\exists\delta>0$ s.t., $\forall x\in\D_{f\cdot g}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,(f\cdot g)_x>M\,)$.\\
Given $M\in\R$. \quad {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_{f\cdot g}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,(f\cdot g)_x>M\,)$.\\
Let $N:=-\sqrt{\max\{M,0\}}$.\\
Then $N\in\R$ and $N\le0$ and $N^2=\max\{M,0\}$.\\
Since \quad as $x\to a$, $f_x\to-\infty$, \quad choose $\alpha>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,0<|x-a|<\alpha\,)\,\Rightarrow\,(\,f_x<N\,)$.\\
Since \quad as $x\to a$, $g_x\to-\infty$, \quad choose $\beta>0$ s.t., $\forall x\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,0<|x-a|<\beta\,)\,\Rightarrow\,(\,g_x<N\,)$.\\
Let $\delta:=\min\{\alpha,\beta\}$. \quad Then $\delta\le\alpha$ and $\delta\le\beta$ and $\delta>0$.\\
{\bf Want:} $\forall x\in\D_{f\cdot g}$, $\,$ $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,(f\cdot g)_x>M\,)$.\\
Given $x\in\D_{f\cdot g}$. \quad Assume $0<|x-a|<\delta$. \quad {\bf Want:} $(f\cdot g)_x>M$.\\
Since $x\in\D_{f\cdot g}$, we get: \quad $x\in\D_f$ \quad and \quad $x\in D_g$.\\
Since $0<|x-a|$ and $|x-a|<\delta\le\alpha$, we get $0<|x-a|<\alpha$.\\
Then, by choice of $\alpha$, we get $f_x<N$. \quad Then $-f_x>-N$.\\
Since $0<|x-a|$ and $|x-a|<\delta\le\beta$, we get $0<|x-a|<\beta$.\\
Then, by choice of $\beta$, we get $g_x<N$. \quad Then $-g_x>-N$.\\
Since $N\le0$, we get $-N\ge0$.\\
Since $-f_x>-N\ge0$ and $-g_x\ge-N\ge0$,\\
\phx\qquad we get $(-f_x)\cdot(-g_x)>(-N)\cdot(-N)$. \qquad
Then $f_x\cdot g_x>N^2$.\\
Then $(f\cdot g)_x=f_x\cdot g_x>N^2=\max\{M,0\}\ge M$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
7-5. Show: Let $f,g:\R\drta\R$, $q\in\D_{f\cdot g}$.\\
Then: $(f\cdot g)_q^\T\,\,=\,\,f_q^\T\cdot g_q\,\,+\,\,f_q\cdot g_q^\T\,\,+\,\,f_q^\T\cdot g_q^\T$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall h\in\R$, \quad
$((f\cdot g)_q^\T)_h\,=\,(f_q^\T\cdot g_q\,\,+\,\,f_q\cdot g_q^\T\,\,+\,\,f_q^\T\cdot g_q^\T)_h$.\\
Given $h\in\R$. \quad
{\bf Want:} $((f\cdot g)_q^\T)_h\,=\,(f_q^\T\cdot g_q\,\,+\,\,f_q\cdot g_q^\T\,\,+\,\,f_q^\T\cdot g_q^\T)_h$.\\
We have $((f\cdot g)_q^\T)_h\,=\,(f\cdot g)_{q+h}-(f\cdot g)_q\,=\,f_{q+h}\cdot g_{q+h}-f_q\cdot g_q$.\\
{\bf Want:} $f_{q+h}\cdot g_{q+h}-f_q\cdot g_q\,=\,(f_q^\T\cdot g_q\,\,+\,\,f_q\cdot g_q^\T\,\,+\,\,f_q^\T\cdot g_q^\T)_h$.\\
Let \quad $a:=f_q$, \quad $b:=g_q$, \quad $A:=f_{q+h}$, \quad $B:=g_{q+h}$.\\
{\bf Want:} $A\cdot B-a\cdot b\,=\,(f_q^\T\cdot g_q\,\,+\,\,f_q\cdot g_q^\T\,\,+\,\,f_q^\T\cdot g_q^\T)_h$.\\
We have $A-a\,=\,f_{q+h}-f_q\,=\,(f_q^\T)_h$.\\
We have $B-b\,=\,g_{q+h}-g_q\,=\,(g_q^\T)_h$.\\
By the Naive Product Rule, we have\\
\phx\qquad $A\cdot B-a\cdot b\,=\,(A-a)\cdot b+a\cdot(B-b)+(A-a)\cdot(B-b)$.\\
Then $A\cdot B-a\cdot b\,=\,(f_q^\T)_h\cdot g_q+f_q\cdot(g_q^\T)_h+(f_q^\T)_h\cdot(g_q^\T)_h$\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,=\,(f_q^\T\cdot g_q)_h+(f_q\cdot g_q^\T)_h+(f_q^\T\cdot g_q^\T)_h$\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,=\,(f_q^\T\cdot g_q+f_q\cdot g_q^\T+f_q^\T\cdot g_q^\T)_h$. \qquad\qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 6: Due on Tuesday 25 October}
\vskip.1in\hrule\vskip.1in\noindent
6-1. Show: $\forall a\in\R$, $\exists x\in\R$ s.t.~$x^5+x^3+x=a$.
\vskip.1in\noindent
{\it Proof:}
Given $a\in\R$. \quad {\bf Want:} $\exists x\in\R$ s.t.~$x^5+x^3+x=a$.\\
We have $-|a|\le a$ and $|a|\ge a$. \quad Let $b:=|a|$.\\
Then $-b\le a$ and $b\ge a$.\\
Also, $b\ge 0$, so $b^5+b^3\ge 0$.\\
Then $-(b^5+b^3)-b\le-b$ and $(b^5+b^3)+b\ge b$.\\
Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^5+x^3+x$. \quad
Then $f$ is continuous.\\
We have $f_{-b}=-(b^5+b^3)-b\le-b\le a$ and $f_b=(b^5+b^3)+b\ge b\ge a$.\\
Then $f_{-b}\le a$ and $f_b\ge a$, and so $f_{-b}\le a\le f_b$, and so $a\in[f_{-b}|f_b]$.\\
Since $f$ is continuous on $[-b|b]$, by the IVT, we have $[f_{-b}|f_b]\subseteq f_*[-b|b]$.\\
Since $a\in[f_{-b}|f_b]\subseteq f_*[-b|b]$ choose $x\in[-b|b]$ s.t.~$f_x=a$.\\
Then $x\in[-b|b]\subseteq\R$. \quad Want $x^5+x^3+x=a$.\\
We have $x^5+x^3+x=f_x=a$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-2. Show: Let $f,g:\R\drta\R$, $a,b\in\R$.\\
Assume: as $x\to a$, $f_x\to b$. \qquad
Assume: $g$ is continuous at $b$.\\
\phx\qquad\qquad Then: as $x\to a$, $(g\circ f)_x\to g_b$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,|(g\circ f)_x-g_b|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,|(g\circ f)_x-g_b|<\vep\,)$.\\
Since $g$ is continuous at $b$, choose $\eta>0$ s.t., $\forall y\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|y-b|<\eta\,)\,\Rightarrow\,(\,|g_y-g_b|<\vep\,)$.\\
Since as $x\to a$, $f_x\to b$, choose $\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-b|<\eta\,)$.\\
Then $\delta>0$. {\bf Want:} $\forall x\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,0<|x-a|<\delta\,)\,\Rightarrow\,(\,|(g\circ f)_x-g_b|<\vep\,)$.\\
Given $x\in\D_{g\circ f}$. \quad Assume $0<|x-a|<\delta$. \quad {\bf Want:} $|(g\circ f)_x-g_b|<\vep$.\\
Since $x\in\D_{g\circ f}$, we get $f_x\in\D_g$. \quad Then $f_x\ne\frownie$, so $x\in\D_f$.\\
Since $0<|x-a|<\delta$, by choice of $\delta$, we get $|f_x-b|<\eta$.\\
Let $y:=f_x$. Then $y=f_x\in\D_g$.\\
Since $|y-b|=|f_x-b|<\eta$, by choice of $\eta$, we get $|g_y-g_b|<\vep$.\\
Then $|(g\circ f)_x-g_b|=|g_{f_x}-g_b|=|g_y-g_b|<\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-3. Show: Let $f:\R\drta\R$, $a,b\in\D_f$.\\
Assume $a\ne b$. \qquad Let $m:=\DQ_f(a,b)$.\\
Define $g:\R\drta\R$ by: \qquad $\forall x\in\R$, \quad $g_x\,=\!\!\!\!\!{}^*\,\,\,f_x-mx$.\\
Then: $g_a=g_b$.
\vskip.1in\noindent
{\it Proof:}
We have $\ds{m=\DQ_f(a,b)=\frac{f_b-f_a}{b-a}}$, so $\ds{m=\frac{f_b-f_a}{b-a}}$.\\
Then $m\cdot(b-a)=f_b-f_a$.\\
Then $mb-ma=f_b-f_a$.\\
Then $f_a-ma=f_b-mb$.\\
Then $g_a=g_b$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-4. Show: Let $f:\R\drta\R$, $a\in\D_f$.\\
Assume $f_a^\T$ is continuous at $0$. \quad Then $f$ is continuous at $a$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Let $g:=f_a^\T$. \quad Then, by assumption, $g$ is continuous at $0$.\\
Since $g$ is continuous at $0$, choose $\delta>0$ s.t., $\forall h\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|h-0|<\delta\,)\,\Rightarrow\,(\,|g_h-g_0|<\vep\,)$.\\
Then $\delta>0$. {\bf Want:} $\forall x\in\D_f$, $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Given $x\in\D_f$. \quad Assume $|x-a|<\delta$. \quad {\bf Want:} $|f_x-f_a|<\vep$.\\
Since $x,a\in\D_f$, it follows that $f_x-f_a\ne\frownie$. \quad Let $h:=x-a$.\\
We have $g_h\,=\,(f_a^\T)_h\,=\,f_{a+h}-f_a\,=\,f_x-f_a$.\\
So, since $f_x-f_a\ne\frownie$, we see that $g_h\ne\frownie$ and $g_h=f_x-f_a$.\\
Since $g_h\ne\frownie$, we get $h\in\D_g$.\\
So, since $|h-0|=|h|=|x-a|<\delta$, by choice of $\delta$, we get $|g_h-g_0|<\vep$.\\
Since $a\in\D_f$, we get $f_{a+0}-f_a=0$.\\
Then $g_0\,\,\,\,=\!\!\!\!\!{}^*\,\,\,\,\,(f_a^\T)_0\,\,=\,f_{a+0}-f_a\,\,\,=\,\,\,0$. \quad Then $g_0=0$.\\
Then $f_x-f_a=g_h=g_h-g_0$. \quad Then $|f_x-f_a|=|g_h-g_0|<\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
6-5. Show: Let $f,g:\R\drta\R$, $a\in\D_{g\circ f}$.
Then $(g\circ f)_a^\T\,=\,(g_{f_a}^\T)\,\circ\,(f_a^\T)$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall h\in\R$, $\,\,$ $((g\circ f)_a^\T)_h\,=\,((g_{f_a}^\T)\,\circ\,(f_a^\T))_h$.\\
Given $h\in\R$. \quad {\bf Want:} $((g\circ f)_a^\T)_h\,=\,((g_{f_a}^\T)\,\circ\,(f_a^\T))_h$.\\
We have: \quad $((g\circ f)_a^\T)_h\,=\,(g\circ f)_{a+h}\,-\,(g\circ f)_a$.\\
{\bf Want:} $(g\circ f)_{a+h}\,-\,(g\circ f)_a\,=\,((g_{f_a}^\T)\,\circ\,(f_a^\T))_h$.\\
Let $k:=((f_a)^\T)_h$. \quad Then: \quad $((g_{f_a}^\T)\,\circ\,(f_a^\T))_h\,=\,(g_{f_a}^\T)_k$.\\
{\bf Want:} $(g\circ f)_{a+h}\,-\,(g\circ f)_a\,=\,(g_{f_a}^\T)_k$.\\
We have: \quad $(g_{f_a}^\T)_k\,=\,g_{f_a+k}\,-\,g_{f_a}$.\\
{\bf Want:} $(g\circ f)_{a+h}\,-\,(g\circ f)_a\,=\,g_{f_a+k}\,-\,g_{f_a}$.\\
We have: \quad $k\,=\,(f_a^\T)_h\,=\,f_{a+h}-f_a$.\\
Then $f_a+k\,=\,f_a+f_{a+h}-f_a$.\\
Since $a\in\D_f$, we get $f_a-f_a=0$, so $f_a+f_{a+h}-f_a\,=\,f_{a+h}$.\\
Then $f_a+k\,=\,f_a+f_{a+h}-f_a\,=\,f_{a+h}$,\\
\phx\qquad so \quad $f_a+k\,=\,f_{a+h}$, \quad so \quad $g_{f_a+k}\,=\,g_{f_{a+h}}$.\\
Then $(g\circ f)_{a+h}\,-\,(g\circ f)_a\,=\,g_{f_{a+h}}\,-\,g_{f_a}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,=\!\!\!\!\!{}^*\,\,\,\,\,\,g_{f_a+k}\,-\,g_{f_a}$. \qquad\qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 5: Due on Wednesday 19 October}
\vskip.1in\hrule\vskip.1in\noindent
5-1. Show: Let $f,g:\R\drta\R$, $a\in\R$.\\
Assume $f$ and $g$ are both continuous at $a$.\\
Then $f+g$ is continuous at $a$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_{f+g}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|(f+g)_x-(f+g)_a|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_{f+g}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|(f+g)_x-(f+g)_a|<\vep\,)$.\\
Since $f$ is continuous at $a$, choose $\lambda>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep/2\,)$.\\
Since $g$ is continuous at $a$, choose $\mu>0$ s.t., $\forall x\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|g_x-g_a|<\vep/2\,)$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \qquad Then $\delta>0$.\\
{\bf Want:} $\forall x\in\D_{f+g}$, \quad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|(f+g)_x-(f+g)_a|<\vep\,)$.\\
Given $x\in\D_{f+g}$. $\,$ Assume $|x-a|<\delta$. $\,$ {\bf Want:} $|(f+g)_x-(f+g)_a|<\vep$.\\
Since $x\in\D_{f+g}$, we get: $x\in\D_f$ and $x\in\D_g$.\\
Since $\delta=\min\{\lambda,\mu\}$, we get: $\delta\le\lambda$ and $\lambda\le\mu$.\\
Since $|x-a|<\delta\le\lambda$, by choice of $\lambda$, we get: $|f_x-f_a|<\vep/2$.\\
Since $|x-a|<\delta\le\mu$, by choice of $\mu$, we get: $|g_x-g_a|<\vep/2$.\\
Then $\,\,|(f+g)_x-(f+g)_a|\,\,=\,\,|(f_x+g_x)-(f_a+g_a)|$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $=\,\,|f_x+g_x-f_a-g_a|$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $=\,\,|f_x-f_a+g_x-g_a|$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\le\,\,|f_x-f_a|+|g_x-g_a|$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $<\,\,(\vep/2)+(\vep/2)\,\,=\vep$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
5-2. Show: Let $f,g:\R\drta\R$, $a\in\R$.\\
Assume \quad $f$ is continuous at $a$ \quad and \quad $g$ is continuous at $f_a$.\\
Then $g\circ f$ is continuous at $a$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|(g\circ f)_x-(g\circ f)_a|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|(g\circ f)_x-(g\circ f)_a|<\vep\,)$.\\
Since $g$ is continuous at $f_a$, choose $\eta>0$ s.t., $\forall y\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|y-f_a|<\eta\,)\,\Rightarrow\,(\,|g_y-g_{f_a}|<\vep\,)$.\\
Since $f$ is continuous at $a$, choose $\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\eta\,)$.\\
Then $\delta>0$.\\
{\bf Want:} $\forall x\in\D_{g\circ f}$, \quad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|(g\circ f)_x-(g\circ f)_a|<\vep\,)$.\\
Given $x\in\D_{g\circ f}$. Assume $|x-a|<\delta$. {\bf Want:} $|(g\circ f)_x-(g\circ f)_a|<\vep$.\\
Since $x\in\D_{g\circ f}$, we get $f_x\in\D_g$, and so $x\in\D_f$.\\
Since $|x-a|<\delta$, by choice of $\delta$, we get $|f_x-f_a|<\eta$.\\
Let $y:=f_x$. \quad Then $|y-f_a|<\eta$. Also, $y=f_x\in\D_g$, so $y\in\D_g$.\\
Since $|y-f_a|<\eta$, by choice of $\eta$, we get $|g_y-g_{f_a}|<\vep$.\\
Then $|(g\circ f)_x-(g\circ f)_a|=|g_{f_x}-g_{f_a}|=|g_y-g_{f_a}|<\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
5-3. Show: Let $f:\R\drta\R$, $s\in(\D_f)^\N$, $a\in\R$.\\
Assume \quad $f$ is continuous at $a$ \quad and \quad $s\to a$. \qquad Then $f\circ s\to f_a$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(f\circ s)_j-f_a|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(f\circ s)_j-f_a|<\vep\,)$.\\
Since $f$ is continuous at $a$, choose $\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Since $s\to a$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|s_j-a|<\delta\,)$.\\
Then $K\in\N$.\\
{\bf Want:} $\forall j\in\N$, \quad $(\,j\ge K\,)\,\Rightarrow\,(\,|(f\circ s)_j-f_a|<\vep\,)$.\\
Given $j\in\N$. Assume $j\ge K$. {\bf Want:} $|(f\circ s)_j-f_a|<\vep$.\\
Since $j\ge K$, by choice of $K$, we get: $|s_j-a|<\delta$.\\
Since $s\in(\D_f)^\N$, we get: $s_j\in\D_f$. Let $x:=s_j$. Then $x\in\D_f$.\\
Since $|x-a|=|s_j-a|<\delta$, by choice of $\delta$, we get: $|f_x-f_a|<\vep$.\\
Then $|(f\circ s)_j-f_a|=|f_{s_j}-f_a|=|f_x-f_a|<\vep$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
5-4. Show: let $f:\R\drta\R$.\\
Assume $f$ is Lipschitz. \quad Then $f$ is uniformly continuous.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w,x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-w|<\delta\,)\,\Rightarrow\,(\,|f_x-f_w|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall w,x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-w|<\delta\,)\,\Rightarrow\,(\,|f_x-f_w|<\vep\,)$.\\
Since $f$ is Lipschitz, choose $L\ge0$ s.t.~$f$ is $L$-Lipschitz.\\
Let $\delta:=\vep\,/\,(L+1)$. \qquad Then $\delta>0$.\\
{\bf Want:} $\forall w,x\in\D_f$, \quad $(\,|x-w|<\delta\,)\,\Rightarrow\,(\,|f_x-f_w|<\vep\,)$.\\
{\bf Want:} Given $w,x\in\D_f$. Assume $|x-w|<\delta$. {\bf Want:} $|f_x-f_w|<\vep$.\\
Since $f$ is $L$-Lipschitz, we get: $|f_x-f_w|\le L\cdot|w-x|$.\\
Since $L\ge0$ and $|w-x|<\delta$, we get $L\cdot|x-w|\le L\cdot\delta$.\\
Then $|f_x-f_w|\le L\cdot|w-x|\le L\cdot\delta=L\cdot\vep\,/\,(L+1)<\vep$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
5-5. Show: let $f:\R\drta\R$.\\
Assume $f$ is uniformly continuous. \quad Then $f$ is continuous.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall a\in\R$, $f$ is continuous at $a$.\\
Given $a\in\R$. \quad {\bf Want:} $f$ is continuous at $a$.\\
{\bf Want:} $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Given $\vep>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Since $f$ is uniformly continuous, choose $\delta>0$ s.t., $\forall w,x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,|x-w|<\delta\,)\,\Rightarrow\,(\,|f_x-f_w|<\vep\,)$.\\
Then $\delta>0$.\\
{\bf Want:} $\forall x\in\D_f$, \quad $(\,|x-a|<\delta\,)\,\Rightarrow\,(\,|f_x-f_a|<\vep\,)$.\\
Given $x\in\D_f$. Assume $|x-a|<\delta$. {\bf Want:} $|f_x-f_a|<\vep$.\\
Let $w:=a$. Then $w=a\in\D_f$. Recall that $x\in\D_f$.\\
Also, $|x-w|=|x-a|<\delta$, so $|x-w|<\delta$.\\
Since $|x-w|<\delta$, by choice of $\delta$, we get: $|f_x-f_w|<\vep$.\\
Then $|f_x-f_a|=|f_x-f_w|<\vep$. \qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 4: Due on Tuesday 11 October}
\vskip.1in\hrule\vskip.1in\noindent
4-1. Show: $(1,2,3,\ldots)\to\infty$.
\vskip.1in\noindent
{\it Proof:}
Let $s:=(1,2,3,\ldots)$. \quad {\bf Want:} $s\to\infty$.\\
We have: $\forall j\in\N$, $s_j=j$.\\
{\bf Want:} $\forall M\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$, \quad $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>M\,)$.\\
Given $M\in\R$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$, \quad $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>M\,)$.\\
By the AP, choose $K\in\N$ s.t.~$K>M$. \quad Then $K\in\N$.\\
{\bf Want:} $\forall j\in\N$, \quad $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>M\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $s_j>M$.\\
We have $s_j=j\ge K>M$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-2. Show: Let $s\in\R^\N$, $c<0$.\\
Assume $s\to\infty$. \qquad Then $c\cdot s\to-\infty$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall N\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,(c\cdot s)_j<N\,)$.\\
Given $N\in\R$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,(c\cdot s)_j<N\,)$.\\
Let $M:=N/c$. \quad Then $M\in\R$ and $c\cdot M=N$.\\
Since $s\to\infty$, choose $K\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>M\,)$.\\
Then $K\in\N$. \qquad {\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,(c\cdot s)_j<N\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $(c\cdot s)_j<N$.\\
Since $j\ge K$, by choice of $K$, we have $s_j>M$.\\
So, \quad since $c<0$, \quad we get $c\cdot s_j<c\cdot M$.\\
Then $(c\cdot s)_j=c\cdot s_j<c\cdot M=N$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-3. Show: Let $s,t\in\R^\N$.\\
Assume $s\to\infty$ and $t\to\infty$. \qquad Then $s+t\to\infty$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall M\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,(s+t)_j>M\,)$.\\
Given $M\in\R$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,(s+t)_j>M\,)$.\\
Since $s\to\infty$, choose $A\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge A\,)\,\Rightarrow\,(\,s_j>M/2\,)$.\\
Since $t\to\infty$, choose $B\in\N$ s.t., $\forall j\in\N$, $\,$ $(\,j\ge B\,)\,\Rightarrow\,(\,t_j>M/2\,)$.\\
Let $K:=\max\{A,B\}$. \qquad Then $K\in\{A,B\}\subseteq\N$.\\
{\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,(s+t)_j>M\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $(s+t)_j>M$.\\
We have $K\ge A$ and $K\ge B$.\\
Since $j\ge K\ge A$, by choice of $A$, we get $s_j>M/2$.\\
Since $j\ge K\ge B$, by choice of $B$, we get $t_j>M/2$.\\
Then $(s+t)_j=s_j+t_j>(M/2)+(M/2)=M$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-4. Show: Let $s\in(\R_0^\times)^\N$.\\
Assume $s\to\infty$. \qquad Then $1/s\to0$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\varepsilon>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(1/s)_j-0|<\varepsilon\,)$.\\
Given $\varepsilon>0$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(1/s)_j-0|<\varepsilon\,)$.\\
Let $M:=1/\varepsilon$. \quad Since $\varepsilon>0$, we get $M\in\R$ and $M>0$ and $1/M=\varepsilon$.\\
Since $s\to\infty$, choose $K\in\N$ s.t., $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,s_j>M\,)$\\
Then $K\in\N$. \quad {\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,|(1/s)_j-0|<\varepsilon\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $|(1/s)_j-0|<\varepsilon$.\\
Since $j\ge K$, by choice of $K$, we have $s_j>M$.\\
We have $s_j>M>0$, so $s_j>0$, so $1/s_j>0$, so $|1/s_j|=1/s_j$.\\
Since $s_j>M>0$, we also get $1/s_j<1/M$.\\
Then $|(1/s)_j-0|=|(1/s)_j|=|1/s_j|=1/s_j<1/M=\varepsilon$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
4-5. Show: Let $s\in(\R_0^\times)^\N$, $c\in\R$.\\
Assume $s\to-\infty$. \qquad Then $c/s\to0$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $\forall\varepsilon>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(c/s)_j-0|<\varepsilon\,)$.\\
Given $\varepsilon>0$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,|(c/s)_j-0|<\varepsilon\,)$.\\
Let $M:=(|c|+1)/\varepsilon$. \quad Then $M\in\R$ and $M>0$.\\
Since $s\to-\infty$, choose $K\in\N$ s.t., $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,s_j<-M\,)$.\\
Then $K\in\N$. \quad {\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,|(c/s)_j-0|<\varepsilon\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $|(c/s)_j-0|<\varepsilon$.\\
Since $j\ge K$, by choice of $K$, we get $s_j<-M$.\\
Since $s_j<-M<0$, it follows that $s_j<0$, so $|s_j|=-s_j$.\\
Since $-s_j>M>0$, we get $1/(-s_j)<1/M$. Then $|c|/(-s_j)\le|c|/M$.\\
We have $1/M=\varepsilon/(|c|+1)$, and so $|c|\cdot(1/M)<\varepsilon$.\\
Then $|c|/|s_j|=|c|/(-s_j)\le|c|/M=|c|\cdot(1/M)<\varepsilon$, so $|c|/|s_j|<\varepsilon$.\\
Then $|(c/s)_j-0|=|(c/s)_j|=|c/s_j|=|c|/|s_j|<\varepsilon$. \qquad
QED
\vskip.1in\noindent
Below is an alternate proof.
\vskip.1in\noindent
{\it Proof:}\\
\\
{\it Claim:} $-s\to\infty$.\\
{\it Proof of Claim:} {\bf Want:} $\forall M\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,(-s)_j>M\,)$.\\
Given $M\in\R$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,(-s)_j>M\,)$.\\
Since $s\to-\infty$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,s_j<-M\,)$.\\
Then $K\in\N$. \qquad {\bf Want:} $\forall j\in\N$, $\,$ $(\,j\ge K\,)\,\Rightarrow\,(\,(-s)_j>M\,)$.\\
Given $j\in\N$. \quad Assume $j\ge K$. \quad {\bf Want:} $(-s)_j>M$.\\
Since $j\ge K$, by choice of $K$, we get: $s_j<-M$. \quad Then $-s_j>M$.\\
Then $(-s)_j=-s_j>M$.\\
{\it End of proof of Claim.}\\
\\
Let $b:=-c$, $r:=-s$. \quad By the Claim, $r\to\infty$.\\
So, by HW 4-4, $1/r\to0$. \quad
So, by HW 4-2, $b\cdot(1/r)\to b\cdot0$.\\
So, since $b\cdot0=0$, we get: \quad $b\cdot(1/r)\to0$. \qquad Want: $c/s\to0$.\\
{\bf Want:} $b\cdot(1/r)=c/s$. \quad
{\bf Want:} $\forall j\in\N$, $(b\cdot(1/r))_j=(c/s)_j$.\\
Given $j\in\N$. \quad {\bf Want:} $(b\cdot(1/r))_j=(c/s)_j$.\\
We have $r_j=(-s)_j=-s_j$, \quad so $r_j=-s_j$. \quad Also, $(c/s)_j=c/s_j$.\\
Then $b/r_j=(-c)/(-s_j)=c/s_j=(c/s)_j$, \quad so $b/r_j=(c/s)_j$.\\
Then $(b\cdot(1/r))_j=b\cdot(1/r)_j=b\cdot(1/r_j)=b/r_j=(c/s)_j$. \qquad
QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 3: Due on Tuesday 4 October}
\vskip.1in\hrule\vskip.1in\noindent
3-1. Using the $0$-PMI, show: Let $x\in\R$. Then: $\forall j\in\N_0$, $|x^j|=|x|^j$.
\vskip.1in\noindent
{\it Proof:}
Let $S:=\{j\in\N\,|\,|x^j|=|x|^j\}$. \qquad {\bf Want:} $S=\N$.\\
$|x^0|=|1|=1=|x|^0$, so $0\in S$.\\
By PMI, want: $\forall j\in S$, $j+1\in S$.\\
Given $j\in S$. \qquad {\bf Want:} $j+1\in S$.\\
Know: $|x^j|=|x|^j$. \qquad {\bf Want:} $|x^{j+1}|=|x|^{j+1}$.\\
We have: $|x^{j+1}|=|x^j\cdot x|=|x^j|\cdot|x|=|x|^j\cdot|x|=|x|^{j+1}$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-2. Show: $\forall\varepsilon>0$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,|x-4|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|x^3-5x^2+4x|<\varepsilon\,\,)$.
\vskip.1in\noindent
{\it Proof:}\\
Given $\varepsilon>0$. {\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,|x-4|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|x^3-5x^2+4x|<\varepsilon\,\,)$.\\
Let $\delta:=\min\{1,\varepsilon/31,\varepsilon/30\}$. \qquad Then $\delta>0$.\\
{\bf Want:} $\forall x\in\R$,\qquad $(\,\,|x-4|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|x^3-5x^2+4x|<\varepsilon\,\,)$.\\
Given $x\in\R$. Assume $|x-4|<\delta$. {\bf Want:} $|x^3-5x^2+4x|<\varepsilon$.\\
We have $\delta\le1$ and $\delta\le\varepsilon/31$.\\
Since $|x|=|x-4+4|\le|x-4|+|4|<\delta+4\le5$, we get $|x|^2+|x|<30$.\\
Then $|x^2-x|\le|x|^2+|x|<30$.\\
Then $|(x-4)\cdot(x^2-x)|=|x-4|\cdot|x^2-x|\le\delta\cdot30\le(\varepsilon/31)\cdot30<\varepsilon$.\\
Then $|x^3-5x^2+4x|=|(x-4)\cdot(x^2-x)|<\varepsilon$. \qquad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-3. Show: Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^2$.\\
\phx\quad $\,$ Then: $\forall a\in\R$, $\forall\varepsilon>0$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,|x-a|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x-f_a|<\varepsilon\,\,)$.
\vskip.1in\noindent
{\it Proof:}
Given $a\in\R$, $\varepsilon>0$.\\
{\bf Want:} $\exists\delta>0$ s.t., $\forall x\in\R$, \quad $(\,\,|x-a|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x-f_a|<\varepsilon\,\,)$.\\
Let $\delta:=\varepsilon/(1+2\cdot|a|)$. \quad Then $\delta>0$.\\
{\bf Want:} $\forall x\in\R$, \qquad $(\,\,|x-a|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x-f_a|<\varepsilon\,\,)$.\\
Given $x\in\R$. Assume $|x-a|<\delta$. {\bf Want:} $|f_x-f_a|<\varepsilon$.\\
We have $|x|=|x-a+a|\le|x-a|+|a|<\delta+|a|$, so $|x|<\delta+|a|$.\\
Then $|x|+|a|<\delta+2\cdot|a|\le1+2\cdot|a|$.\\
Then $\delta\cdot(|x|+|a|)=(\varepsilon/(1+2\cdot|a|))\cdot(|x|+|a|)$\\
\phx\qquad\qquad\qquad\qquad $\le(\varepsilon/(1+2\cdot|a|))\cdot(1+2\cdot|a|)<\varepsilon$.\\
Then $|(x-a)\cdot(x+a)|=|x-a|\cdot|x+a|\le\delta\cdot|x+a|\le\delta\cdot(|x|+|a|)<\varepsilon$.\\
Then $|f_x-f_a|=|x^2-a^2|=|(x-a)\cdot(x+a)|<\varepsilon$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
3-4. Show: $\forall M\in\R$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\quad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,j^4-7j-9>M\,\,)$.
\vskip.1in\noindent
{\it Proof:}
Given $M\in\R$. {\bf Want:} $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\quad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,j^4-7j-9>M\,\,)$.\\
Let $L:=\max\{2,M+9\}$. \quad Then \quad $L\le2$ \quad and \quad $L\le M+9$.\\
By the Archimedean Principle,\\
\phx\quad choose $K\in\N$ such that $K>L$. \qquad Then $K\in\N$.\\
{\bf Want:} $\forall j\in\N$, \qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,j^4-7j-9>M\,\,)$.\\
Given $j\in\N$. Assume $j\ge K$. {\bf Want:} $j^4-7j-9>M$.\\
We have $j-9\ge K-9>L-9\ge(M+9)-9=M$, so $j-9>M$.\\
We have $j\ge K>L\ge2$. \quad Then $j^3>8$.\\
So, since $j\in\N>0$, we get $j^3\cdot j>8j$.\\
Then $j^4-7j-9=j^3\cdot j-7j-9>8j-7j-9=j-9>M$. \quad QED
\vskip.1in\hrule\vskip.1in\noindent
3-5. Show: \qquad Let $x,y\in\R$. \quad Then: $\,\,$ $|\,\,|y|\,\,-\,\,|x|\,\,|\,\,\,\,\le\,\,\,\,|\,\,y\,\,-\,\,x\,\,|$.
\vskip.1in\noindent
{\it Proof:}
Let $a:=|x|$ and $b:=|y|$ and $\varepsilon:=|y-x|$. \qquad {\bf Want:} $|b-a|\le\varepsilon$.\\
{\bf Want:} $a-\varepsilon\le b\le a+\varepsilon$.\\
{\bf Want:} \quad (1) $a-\varepsilon\le b$ \quad and \quad (2) $b\le a+\varepsilon$.\\
\\
{\it Proof of (1):}\\
We have $a=|x|=|y+(x-y)|\le|y|+|x-y|=|y|+|y-x|=b+\varepsilon$.\\
Then $a\le b+\varepsilon$. \quad Then $a-\varepsilon\le b$.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
We have $b=|y|=|x+(y-x)|\le|x|+|y-x|=a+\varepsilon$.\\
Then $b\le a+\varepsilon$.\\
{\it End of proof of (2).} \qquad QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 2: Due on Tuesday 27 September}
\vskip.1in\hrule\vskip.1in\noindent
2-1. Show: Let $S\subseteq\R^*$. Then: $\sup_S\,\,\,\,=^*\,\,\max_S$.
\vskip.1in\noindent
{\it Proof:}
{\bf Want:} $(\,\,\max_S\ne\frownie\,\,)\,\,\Rightarrow\,\,(\,\,\sup_S=\max_S\,\,)$.\\
Assume $\max_S\ne\frownie$. \quad {\bf Want:} $\sup_S=\max_S$.\\
We have: $\max_S\,\,=\,\,\UE(S\cap\UB_S)\,\,\,{}^*\!\!\in\,\,S\cap\UB_S$, so $\max_S\in S\cap\UB_S$.\\
Since $\max_S\in S\le\sup_S$, we get $\max_S\le\sup_S$.\\
Since $\max_S\in\UB_S\ge\sup_S$, we get $\max_S\ge\sup_S$.\\
Since $\max_S\le\sup_S$ and $\max_S\ge\sup_S$,\\
\phx\qquad we get $\max_S=\sup_S$, \quad so $\sup_S=\max_S$.
\quad QED
\vskip.1in\hrule\vskip.1in\noindent
2-2. Show: $\forall M\in\R$, $\exists N\in\R$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad $(\,\,x<N\,\,)\,\,\Rightarrow\,\,(\,\,x^2>M\,\,)$.
\vskip.1in\noindent
{\it Proof:}
Given $M\in\R$. {\bf Want:} $\exists N\in\R$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,x<N\,\,)\,\,\Rightarrow\,\,(\,\,x^2>M\,\,)$.\\
Let $N:=-\sqrt{\max\{M,0\}}$. \quad Then $N\in\R$.\\
{\bf Want:} $\forall x\in\R$, $(\,\,x<N\,\,)\,\,\Rightarrow\,\,(\,\,x^2>M\,\,)$.\\
Given $x\in\R$. Assume $x<N$. {\bf Want:} $x^2>M$.\\
Since $N=-\sqrt{\max\{M,0\}}$, we get $N\le0$ and $N^2=\max\{M,0\}$.\\
Since $x<N\le0$, we get $x^2>N^2$.\\
Then $x^2>N^2=\max\{M,0\}\ge M$.
\quad QED
\vskip.1in\hrule\vskip.1in\noindent
2-3. Show: $\forall M\in\R$, $\exists K\in\N_0$ s.t., $\forall j\in\N_0$,\\
\phx\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,2^j>M\,\,)$.
\vskip.1in\noindent
{\it Proof:}
Given $M\in\R$. {\bf Want:} $\exists K\in\N_0$ s.t., $\forall j\in\N_0$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,2^j>M\,\,)$.\\
By the Archimedean Principle, choose $K\in\N$ such that $K>M$.\\
Then $K\in\N\subseteq\N_0$, so $K\in\N_0$.\\
{\bf Want:} $\forall j\in\N_0$, $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,2^j>M\,\,)$.\\
Given $j\in\N_0$. Assume $j\ge K$. {\bf Want:} $2^j>M$.\\
By an in class theorem, $2^j\ge j+1$.\\
Then $2^j\ge j+1>j\ge K>M$.
\quad QED
\vskip.1in\hrule\vskip.1in\noindent
2-4. Show: $\forall\varepsilon>0$, $\exists K\in\N_0$ s.t., $\forall j\in\N_0$,\\
\phx\qquad\qquad\qquad $\ds{(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,\left(\,\,\frac{1}{2^j}<\varepsilon\,\,\right)}$.
\vskip.1in\noindent
{\it Proof:}
Given $\varepsilon>0$. {\bf Want:} $\exists K\in\N_0$ s.t., $\forall j\in\N_0$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\ds{(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,\left(\,\,\frac{1}{2^j}<\varepsilon\,\,\right)}$.\\
Let $M:=1/\varepsilon$. Then $M\in\R$ and $M>0$ and $\ds{\frac{1}{M}=\varepsilon}$.\\
By HW 2-3, choose $K\in\N_0$ s.t. $\forall j\in\N_0$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,2^j>M\,\,)$.\\
Then $K\in\N_0$. \qquad
{\bf Want:} $\forall j\in\N_0$, $\ds{(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,\left(\,\,\frac{1}{2^j}<\varepsilon\,\,\right)}$.\\
Given $j\in\N_0$. Assume $j\ge K$. {\bf Want:} $\ds{\frac{1}{2^j}<\varepsilon}$.\\
By the choice of $K$, since $j\ge K$, we get $2^j>M$.\\
Since $2^j>M>0$, we get $\ds{\frac{1}{2^j}<\frac{1}{M}}$.\\
Then $\ds{\frac{1}{2^j}<\frac{1}{M}=\varepsilon}$.
\quad QED
\vskip.1in\hrule\vskip.1in\noindent
2-5. Using the $0$-PMI, show: $\forall k\in\N_0$,\\
\phx\qquad\qquad\qquad $1+2+4+8+\cdots+2^k\,\,=\,\,2^{k+1}-1$.
\vskip.1in\noindent
{\it Proof:}
Let $S:=\{j\in\N_0\,|\,1+2+4+8+\cdots+2^k\,\,=\,\,2^{k+1}-1\}$.\\
{\bf Want:} $S=\N_0$. \quad Since $1=2^{0+1}-1$, we get $0\in S$.\\
By the $0$-PMI, want: $\forall j\in S$, $j+1\in S$.\\
Given $j\in S$. \qquad {\bf Want:} $j+1\in S$.\\
Know: $1+2+4+8+\cdots+2^k\,\,=\,\,2^{k+1}-1$.\\
{\bf Want:} $\,1+2+4+8+\cdots+2^k+2^{k+1}\,\,=\,\,2^{(k+1)+1}-1$.\\
We have $\,1+2+4+8+\cdots+2^k+2^{k+1}$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{k+1}-1+2^{k+1}$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{k+1}+2^{k+1}-1$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{k+1}\cdot1+2^{k+1}\cdot1-1$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{k+1}\cdot(1+1)-1$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{k+1}\cdot2-1$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{k+1}\cdot2^1-1$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,\,2^{(k+1)+1}-1$.\\
\quad QED
\vskip.1in\noindent
\hrule\vskip.1in\hrule
\vskip.1in\noindent
\centerline{Homework 1: Due on Tuesday 20 September}
\vskip.1in\hrule\vskip.1in\noindent
1-1. Show: $\forall\varepsilon>0$, $\exists\delta>0$ s.t.~$3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$.
\vskip.1in\noindent
{\it Proof:}
Given $\varepsilon>0$. \quad {\bf Want:} $\exists\delta>0$ s.t.~$3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$.\\
Let $\ds{\delta:=\min\{\,\,\sqrt[8]{2\varepsilon/3}\,\,,\,\,\sqrt[6]{2\varepsilon/7}\,\,,\,\,\sqrt[4]{\varepsilon}\,\,\}}$. \quad
Then $\delta>0$.\\
{\bf Want:} $3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$.\\
Since $0\le\delta\le\sqrt[8]{2\varepsilon/3}$, we get $\delta^8\le(\sqrt[8]{2\varepsilon/3})^8$.\\
Since $0\le\delta\le\sqrt[6]{2\varepsilon/7}$, we get $\delta^6\le(\sqrt[6]{2\varepsilon/7})^6$.\\
Since $0\le\delta\le\sqrt[4]{\varepsilon}$, we get $\delta^4\le(\sqrt[4]{\varepsilon})^4$.\\
Then $\delta^8\le2\varepsilon/3$ and $\delta^6\le2\varepsilon/7$ and $\delta^4\le\varepsilon$.\\
Then $3\delta^8\le2\varepsilon$ and $7\delta^6\le2\varepsilon$ and $2\delta^4\le2\varepsilon$.\\
Then $3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-2. Show: Let $\varepsilon>0$. Then $\exists\delta>0$ s.t.~$3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$.
\vskip.1in\noindent
{\it Proof:}
Let $\ds{\delta:=\min\{\,\,\sqrt[8]{2\varepsilon/3}\,\,,\,\,\sqrt[6]{2\varepsilon/7}\,\,,\,\,\sqrt[4]{\varepsilon}\,\,\}}$. \quad
Then $\delta>0$.\\
{\bf Want:} $3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$.\\
Since $0\le\delta\le\sqrt[8]{2\varepsilon/3}$, we get $\delta^8\le(\sqrt[8]{2\varepsilon/3})^8$.\\
Since $0\le\delta\le\sqrt[6]{2\varepsilon/7}$, we get $\delta^6\le(\sqrt[6]{2\varepsilon/7})^6$.\\
Since $0\le\delta\le\sqrt[4]{\varepsilon}$, we get $\delta^4\le(\sqrt[4]{\varepsilon})^4$.\\
Then $\delta^8\le2\varepsilon/3$ and $\delta^6\le2\varepsilon/7$ and $\delta^4\le\varepsilon$.\\
Then $3\delta^8\le2\varepsilon$ and $7\delta^6\le2\varepsilon$ and $2\delta^4\le2\varepsilon$.\\
Then $3\delta^8+7\delta^6+2\delta^4\le6\varepsilon$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-3. Show: $\forall N\in\R$, $\exists\delta>0$ s.t.~$\forall x\in\R$,\\
\phx\qquad\qquad $(\,-\delta<x<0\,)\,\Rightarrow\,(\,1/x<N\,)$.
\vskip.1in\noindent
{\it Proof:}
Given $N\in\R$. \quad {\bf Want:} $\exists\delta>0$ s.t.~$\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,-\delta<x<0\,)\,\Rightarrow\,(\,1/x<N\,)$.\\
Let $A:=\max\{-N,1\}$. \quad Then $\,$ $A\ge-N$ $\,$ and $\,$ $A\ge1$.\\
We have $A\ge1>0$, \quad so $A>0$.\\
Let $\delta:=1/A$. \quad Then $1/\delta=A$. \quad Also, $\delta>0$.\\
{\bf Want:} $\forall x\in\R$, \quad $(\,-\delta<x<0\,)\,\Rightarrow\,(\,1/x<N\,)$.\\
Given $x\in\R$. \quad Assume $-\delta<x<0$. \quad {\bf Want:} $1/x<N$.\\
Since $-\delta<x<0$, we get $\delta>-x>0$, so $1/\delta<1/(-x)$.\\
Then $-1/\delta>-1/(-x)=1/x$, so $1/x<-1/\delta$.\\
Since $1/\delta=A\ge-N$, we get: $-1/\delta\le N$.\\
Then $1/x<-1/\delta\le N$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-4. Show: Let $N\in\R$. Then $\exists\delta>0$ s.t.~$\forall x\in\R$,\\
\phx\qquad\qquad $(\,-\delta<x<0\,)\,\Rightarrow\,(\,1/x<N\,)$.
\vskip.1in\noindent
{\it Proof:}
Let $A:=\max\{-N,1\}$. \quad Then $\,$ $A\ge-N$ $\,$ and $\,$ $A\ge1$.\\
We have $A\ge1>0$, \quad so $A>0$.\\
Let $\delta:=1/A$. \quad Then $1/\delta=A$. \quad Also, $\delta>0$.\\
{\bf Want:} $\forall x\in\R$, \quad $(\,-\delta<x<0\,)\,\Rightarrow\,(\,1/x<N\,)$.\\
Given $x\in\R$. \quad Assume $-\delta<x<0$. \quad {\bf Want:} $1/x<N$.\\
Since $-\delta<x<0$, we get $\delta>-x>0$, so $1/\delta<1/(-x)$.\\
Then $-1/\delta>-1/(-x)=1/x$, so $1/x<-1/\delta$.\\
Since $1/\delta=A\ge-N$, we get: $-1/\delta\le N$.\\
Then $1/x<-1/\delta\le N$. \quad
QED
\vskip.1in\hrule\vskip.1in\noindent
1-5. Using the PMI, show:\\
\phx\qquad\qquad $\ds{\forall k\in\N,\qquad1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}}$.
\vskip.1in\noindent
{\it Proof:}
Let $\ds{S:=\left\{k\in\N\,\bigg|\,1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}\right\}}$.\\
\\
{\bf Want:} $S=\N$.\qquad
Since $\ds{1^2=\frac{1\cdot(1+1)\cdot(2\cdot1+1}{6}}$, we get $1\in S$.\\
\\
By the PMI, we wish to show: $\forall k\in S$, $k+1\in S$.\\
Given $k\in S$. \qquad {\bf Want:} $k+1\in S$.\\
\\
Know: $\ds{1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}}$.\\
\\
{\bf Want:} $1^2+2^2+\cdots+k^2+(k+1)^2=$\\
\\
\phx\qquad\qquad\qquad $\ds{\frac{(k+1)((k+1)+1)(2\cdot(k+1)+1)}{6}}$.\\
\\
We have
$\ds{1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{k(k+1)(2k+1)}{6}+\frac{6\cdot(k+1)^2}{6}}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{k(k+1)(2k+1)+6\cdot(k+1)^2}{6}}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{(k+1)\cdot(k(2k+1)+6\cdot(k+1))}{6}}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{(k+1)\cdot(2k^2+k+6k+6)}{6}}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{(k+1)\cdot(2k^2+7k+6)}{6}}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{(k+1)\cdot(k+2)\cdot(2k+3)}{6}}$\\
\\
\phx\qquad\qquad\qquad $\ds{=\frac{(k+1)\cdot((k+1)+1)\cdot(2\cdot(k+1)+1)}{6}}$.
\quad QED
\vskip.1in\hrule\vskip.1in\hrule
\end{document}
